Is every self-adjoint operator bounded?

The Closed Graph Theorem is a good way. If $x_n\rightarrow x$ and $Tx_n\rightarrow y$, then $$ \langle y,z\rangle = \lim_n \langle Tx_n,z\rangle=\lim_n \langle x_n,Tz\rangle=\langle x,Tz\rangle = \langle Tx,z\rangle,\;\;\; z\in H. $$ This implies that $Tx=y$. Therefore $T$ has a closed graph. So $T$ is bounded.

This follows from the Uniform Boundedness Principle. $\newcommand{\ip}[2]{\langle #1,#2\rangle}$ Say $B$ is the closed unit ball of $H$. For $x\in H$ define $\Lambda_x:H\to\Bbb C$ by $$\Lambda_x y=\ip{Tx}y.$$Note that each $\Lambda_x$ is bounded, in fact with norm $\|Tx\|$. For every $y\in H$ we have $$\sup_{x\in B}|\Lambda_xy|=\sup_{x\in B}|\ip x{Ty}|=\|Ty\|<\infty.$$So Uniform Boundedness implies that $$\sup_{x\in B}\|Tx\|=\sup_{x\in B}||\Lambda_x||<\infty,$$hence $T$ is bounded.

The operator you has mentioned

$T : H \to H$ a linear operator such that $\forall x,y \in H \; \langle\,Tx,y\rangle=\langle\,x,Ty\rangle $

is an everywhere-defined symmetric operator on a Hilbert space so it is self-adjoint and by Hellinger–Toeplitz theorem it is a bounded operator, but this is not always the situation because there are many self-adjoint operators defined in a dense subspace of $H$ and they are not-bounded.

Note: The self-adjoint operators defined in a dense subspace of $H$, we cannot extend them in an everywhere-defined symmetric operator, we can get only extension for them which is not symmetric the following example is a self-adjoint operator but not bounded.

Let $\Omega$ be any open subset of $ \mathbb{R}^n$ and $(T, D)$ such that $H= L^2(\Omega)$ and $$T= -\Delta \quad and \quad D=\{u\in L^2(\Omega) , \Delta u \in L^2(\Omega)\}$$