this is the question and my attempted answer below. I would be happy if you could please proofread and let me know any suggestions if you spot any errors! :D
Question: Let $H$ be a Hilbert space and $A: H \rightarrow H$ be linear. Assume that $$ (A x, y)=(x, A y) \quad \text { for every } x, y \in H . $$
Show that $A$ is bounded.
My Answer: Let $H$ be a Hilbert space and $A: H \rightarrow H$ be linear. Assume that $$ (A x, y)=(x, A y) \quad \text { for every } x, y \in H $$
We need to show that $A$ is bounded.
To use the Closed Graph Theorem, we will first show that $G(A)$ is closed. This means that if $\left(x_n, A x_n\right)$ converges to $(x, y)$ in $H \times H$, then $(x, y)$ is in $G(A)$, i.e., $y=A x$.
Let $\left(x_n, A x_n\right)$ be a sequence in $G(A)$ that converges to $(x, y)$ in $H \times H$. This implies that $x_n \rightarrow x$ and $A x_n \rightarrow y$ in $H$.
Now, because $H$ is a Hilbert space, it has an inner product that is continuous. Hence, for every $z \in H$, we have $$ \left(A x_n, z\right) \rightarrow(y, z) \text { and }\left(x_n, A z\right) \rightarrow(x, A z) . $$
By the linearity and symmetry of the inner product, and by the given property of $A$, we have $$ \left(A x_n, z\right)=\left(x_n, A z\right) \text { for all } n . $$
Taking the limit as $n \rightarrow \infty$, we get $$ (y, z)=(x, A z) . $$
Since $z$ is arbitrary, we can replace $z$ by $A^* z^{\downarrow}$, where $A^*$ is the adjoint operator of $A$, obtaining $$ \left(y, A^* z\right)=\left(x, A A^* z\right) $$
By the definition of the adjoint, we have $$ (A y, z)=\left(A A^* x, z\right) $$
This implies that $A y=A A^* x$.
Given that the inner product $(x, y)$ is equal to $(A x, A y)$ for every $x$ and $y$, and that $\left(x_n, A x_n\right)$ converges to $(x, y)$, it follows that $(x, y) \in G(A)$, thus $G(A)$ is closed.
Now, by the Closed Graph Theorem, because $G(A)$ is closed, $A$ must be continuous, which in the context of a Hilbert space means that $A$ is bounded.
Hence, we have shown that $A$ is bounded.