Why does this proof work: Closed unit ball in $C_0$ is not compact

Question

I know that this question has been asked to death, and multiple solutions are given, but I still don't understand why the "standard" proof works

Following Show that the closed unit ball $B[0,1]$ in $C[0,1]$ is not compact

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Let $C_o([0,1])$ be the space of continuous functions on $[0,1]$. Define $\|f\| = \sup\{|f(x)| x \in [0,1]]\}$

Let $B[0,1] = \{f \in C_o|\|f\|\leq 1\}$, show that this Closed, Unit ball in $C_o$ is not compact.

Proof Sketch:

Let $\{f_n\} \subset C_o, f_n = x^n, x \in [0,1]$, then $f_n \in C_o, \|f_n\| = 1 \thinspace \forall n$, so $f_n \in B([0,1]) \thinspace \forall n$

But $f_n \to f \notin C_o$. So $B([0,1])$ is not compact.

Three questions:

1. if $f_n \to f \notin C_o$, wouldn't that mean $B([0,1])$ is not closed in the first place (contradicts with assumption)?

2. How does the above implies that no subsequence converges in $B([0,1])$

3. Is the proof sketch essentially correct?

Answer 1

Hint: The proof sketch is correct If a subsequence of $(f_n)$ converges, it converges pointwise towards $f$ which is not continue at 1, so $B(0,1)\subset C([0,1])$ is not compact.

Answer 2

Another exmple to show that $B[0,1]$ is not compact: For $n\in N $ let $0<a_n<b_n<a_{n+1}<1.$ For $x\in [0,a_n]\cup [a_{n+1},1]$ let $f_n(x)=0.$ For $x\in [a_n,b_n]$ let $f_n(x)$ be linear with $f_n(b_n)=1.$ For $x\in [b_n,a_{n+1}]$ let $f_n(x)$ be linear. Then $1=\|f_n\|=\|f_n-f_m\|$ for $m,n\in N$ with $m\ne n$. So $\{f_n:n\in N\}$ is an infinite closed discrete subspace of the space $B[0,1].$ A metric space with an infinite closed discrete subspace is not a compact space.