After figuring out the example given by GEdgar, I realized that I interpreted the set up of the original example totally wrong.
Quote from the original answer:
Let $X_1$ and $X_2$ have the discrete topology. Then $X=X_1\times X_2$ has the discrete topology, so $\mathcal{B}_X=P(X)$. But if $X_1$ and $X_2$ are large enough $\mathcal{B}_{X_1}\otimes\mathcal{B}_{X_2}$ will not be all of $P(X)$. Here's one way to prove this. For $S\subset X$ and $x\in X_1$, let $S_x=\{y\in X_2:(x,y)\in S\}$. Now let $$\mathcal{A}=\{S:|\{S_x:x\in X_1\}|\leq 2^{\aleph_0}\}.$$
Then it is straightforward to check that $\mathcal{A}$ is a $\sigma$-algebra (the key point is that $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$). Since $\mathcal{A}$ contains all rectangles, it follows that $\mathcal{B}_{X_1}\otimes \mathcal{B}_{X_2}\subseteq\mathcal{A}$. But if $|X_1|>2^{\aleph_0}$ and $2^{|X_2|}>2^{\aleph_0}$, then $\mathcal{A}$ is not all of $P(X)$, since there are more than $2^{\aleph_0}$ different choices for $S_x$ and there are enough points $x\in X_1$ to build a set $S$ that realizes more than $2^{\aleph_0}$ of them.
In the collection $\mathcal{A}$, we collect subsets of $X$ that admit at most $2^{\aleph_0}$ many variants of $X_2$ slices (slices in $X$ with fixed first component $x \in X_1$), in the other word, if $S \in \mathcal{A}$, then there are at most $2^{\aleph_0}$ many variants of $S_x$, where $x$ ranges over the whole $X_1$.
Then clearly, the generators of the product $\sigma$-algebra have only $1$ variant of $S_x$. Hence lie the collection $\mathcal{A}$. Then after similar cardinality calculation as below, the sets in the product $\sigma$-algebra can have at most $2^{\aleph_0}$ variants of $S_x$, which implies that $\mathcal{A}$ is itself a $\sigma$-algebra that contains the product $\sigma$-algebra.
However, since there're more then $|X_2|=2^{\aleph_0}$ possible subsets of $X_2$, and we have more than $2^{\aleph_0}$ points in $X_1$, we can find $S \in X$ such that $S$ has more than $2^{\aleph_0}$ variants of $S_x$, which will then lies in $\mathcal{B}(X) \backslash \mathcal{A}$. Therefore
$$
\mathcal{B}(X_1)\otimes\mathcal{B}(X_2) \subset \mathcal{A} \subsetneq \mathcal{B}(X).
$$
This gives another counter-example for the converse of the theorem, which is in a smaller space than the below one. It will still be well appreciated if someone can give some insight on the necessity of separability or on the product space of spaces of cardinality $2^{\aleph_0}$.
$---------------------------------------$
I actually figured out the above one or two days after I posted this answer. I was too lazy to add it in. Since the original question is more important to me (in the sense that it confused me more), I am posting it above the extra example.
Original post:
So after some digging, I find similar question[1] looking for detail to prove that the set $\Delta = \{ (,): \in \}$, given in the post mentioned by GEdgar is the desired counter-example.
Together with the property of the cardinality of $\sigma$-algebra as a consequence of result mentioned in this post[2], we can attempt to prove this assertion.
Let $(X, \mathcal{B}(X))$ be a measurable space with $\mathcal{B}(X)$ being its Borel $\sigma$-algebra and $|X|>2^{\aleph_0}$. Now consider the product $\sigma$-algebra
$$
\mathcal{B}(X)\otimes\mathcal{B}(X)=\sigma(\{\pi^{-1}_{i}(E)| E\in\mathcal{B}(X), \text{i = 1,2}\}),
$$
where $\pi_{i}$ is the projection map from the product space $X\times X$ to its first and second component respectively. $\sigma(A)$ denotes the $\sigma$-algebra generated by the set $A$. And consider the Borel $\sigma$-algebra on the product space $X\times X$, denoted by $\mathcal{B}(X\times X)$.
Then the diagonal
$$
\Delta= \{ (x,x)|x\in X \}
$$
lies in $\mathcal{B}(X\times X) \backslash \mathcal{B}(X)\otimes\mathcal{B}(X)$. Therefore the Borel $\sigma$-algebra of the product space $\mathcal{B}(X\times X)$ strictly contains the product $\sigma$-algebra.
Proof:
This assertion follows from following two properties:
(a) The cardinality of the $\sigma$-algebra of a set of cardinality less or equal to $2^{\aleph_0}$ is at most $2^{\aleph_0}$, i.e. if $|A|\le 2^{\aleph_0}$, then for any $\sigma$-algebra $\mathfrak{m}\in \mathcal{P}(A)$, $|\mathfrak{m}|\le2^{\aleph_0}$.
(b) The diagonal $\Delta$ can only be generated by the singleton sets from the product $\sigma$-algebra.
(a) is the consequence of the fact $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$. If subsets of $A$ has at most $2^{\aleph_0}$ cardinality, then countable union of these subsets (including complement) will have at most $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$ cardinality. By the same process, we can enlarge the collection of subsets every time, but they will all have the same cardinality $2^{\aleph_0}$. However to reach the final $\sigma$-algebra we are looking for, it seems unavoidable to utilize the transfinite induction, which can be found here and in the answer of this post[2]. The conclusion is, we end up with the cardinality $\aleph_1 \cdot 2^{\aleph_0} = 2^{\aleph_0}$.
To prove (b) we notice that the generators of the product $\sigma$-algebra are vertical/horizontal strips in $X\times X$. The intersection of strips can give us a set as union of more than $2^{\aleph_0}$ boxes, but those boxes will share the same first/second coordinate. Since none of pair of two elements in $\Delta$ share a coordinate, sets of this form cannot generate $\Delta$. Thus $\Delta$ shall be generated by sets of single box. In fact, such sets can only contain one element, otherwise it will contain two elements with same coordinate.
Now to prove the assertion, we notice that $|\Delta|=|X|>2^{\aleph_0}$. By (b), if we can generate $\Delta$ we must start with the singleton sets $\{(x,x)\}$, $x\in X$, which has cardinality $1$. Then by (a), the $\sigma$-algebra generated by singleton sets has at most $2^{\aleph_0}$ cardinality, and hence cannot contain $\Delta$.
$$\tag*{$\blacksquare$}$$
As mentioned in the answer of this question[1], there's another proof of this assertion, which can be found here.
This resolved my original question, which was to find an example such that product Borel $\sigma$-algebra is strictly smaller to the Borel $\sigma$-algebra of the product space. But I'm still really confused how can we actually achieve the same goal by the construction in the answer in my original question. I can now see that the "huge" cardinality plays an important role in constructing these examples.
However more questions arise. What happens if we consider spaces with $2^{\aleph_0}$ cardinality and without separability? Will $\mathcal{B}(X\times X) = \mathcal{B}(X)\otimes\mathcal{B}(X)$ hold or not?
