Graph of symmetric linear map is closed

A homework problem:

Let $H$ be a Hilbert space.
Let $T:H\rightarrow H$ be a symmetric linear map ($\langle Tx,y\rangle=\langle x,Ty\rangle$).

Show that $S$ is bounded.

My attempt: I’d like to use the closed graph theorem. I take $(x_n)\subset H$ and assume $x_n \rightarrow x$ and $Tx_n\rightarrow y$. I’d like to show $Tx=y$.
So I calculate:
$\|Tx_n-Tx\|^2=\|T(x_n-x)\|^2=|\langle T(x_n-x), T(x_n-x)\rangle|=$
$|\langle x_n-x, T(T(x_n-x))\rangle|\leq \|x_n-x\|\cdot \|T(T(x_n-x))\|$.

So, it’s enough to show that $\|T(T(x_n-x))\|$ is bounded. The fact that $T(x_n)$ converges tells me that $\|T(x_n-x)\|$ is bounded, but I don’t know what about $\|T(T(x_n-x))\|$.

Solutions Collecting From Web of "Graph of symmetric linear map is closed"

Assume $x=0$ (which is possible by linearity, working if necessary with $x’_n:=x_n-x$, so $Tx’_n\to Tx_n-Tx=y-Tx$), and write
$$\langle y,y\rangle=\lim_{n\to +\infty}\langle Tx_n,y\rangle\underset{\color{red}{\mbox{sym}}}{=}\lim_{n\to +\infty}\langle x_n,Ty\rangle,$$
and conclude using the fact that $x_n\to 0$.

(so actually, in $\lVert y-Tx_n\rVert$, we just need to take the limit at one side of the inner product, not both).