To show that the given bounded linear operator is the zero operator

Question

Let $T:X\to X$ be a bounded linear operator on a complex inner product space X. If $\langle Tx,x=0\ \forall x\in X$, Show that $T=0$.

I didn't use the bounded-ness of T in the proof. I was wondering whether I made an error or is it just given for no reason.

Let $x,y \in X,\ x+iy \in X $

$\implies \langle T(x+iy),(x+iy)\rangle=0\ \forall x,y \in X$

$\implies \langle Tx,x\rangle+i\langle Ty,x\rangle-i\langle Tx,y\rangle+\langle Ty,y\rangle=0\ \forall x,y \in X$

$\implies i\langle Ty,x\rangle-i\langle Tx,y\rangle=0\ \forall x,y \in X$

$\implies \langle Ty,x\rangle=\langle Tx,y\rangle\ \forall x,y \in X$

Also $ \forall x,y \in X, \ x+y \in X$

$\implies \langle T(x+y),x+y\rangle=0\ \forall x,y \in X$

By simplifying we get, $ \langle Tx,y\rangle+\langle Ty,x\rangle=0\ \forall x,y \in X$

Therefore, $ \langle Ty,x\rangle=\langle Tx,y\rangle=0\ \forall x,y \in X$

In Particular let $y=Tx\ \in X$

We have $\langle Tx,Tx\rangle =0\ \forall x\in X$

$\implies T(X)=\{0\}$ or $T=0$