Assume $E$ is a Hilbert space and $A: E \to E$ is a linear operator such that for all $x,y \in E$ $\langle Ax, y \rangle = \langle x, Ay \rangle$.
How can I show, that $A$ is continuous?
My attempt: If it is not bounded ($\Leftrightarrow$ not continuous), then there is a sequence $(x_n)$ in $E$ such that $x_ n \to 0$, but $\|Ax_n\| \to \infty$ (as $n \to \infty$). Furthermore, $(Ax_n, y) = (x_n, Ay) \to (0, Ay) = 0$ which, I suppose, implies, that $Ax_n \rightharpoonup A0 = 0$. A weak convergence in a Hilbert space implies, that $(Ax_n)$ is bounded $\Leftrightarrow \exists M \in \mathbb{R}, \ M > 0, \ \forall n \in \mathbb{N} \ \ ( \|Ax_n\| \leq M )$. This is a contradiction, so $A$ is bounded and hence continuous.
Is it a correct proof?
Personally, your proof seems correct. But perhaps one should say that the adjoint operator and consequently self-adjoint operator it is well defined if you consider $A \in \mathcal{L}(H)$, in other words, is it well defined if we can use the Riesz representation theorem. Even if we consider only normed spaces, to define the transposed/adjoint or the dual operator we need of continuity.
So I think there are so many linear operators characterized by the identity $\langle Ax , y \rangle = \langle x , Ay \rangle$ that are not well defined. Maybe someone more experienced than I can expand my answer.