If a linear operator has an adjoint operator, it is bounded

Show that $T$ and $S$ are closed. Then apply the closed graph theorem to conclude that $T$ and $S$ are continuous.

To show that $T$ is closed, suppose $\{ x_{n} \}$ converges to $x$ and suppose $\{ Tx_{n} \}$ converges to $y$, and show that $Tx=y$; to do this observe that $$ (Tx,z) = (x,Sz) = \lim_{n}(x_{n},Sz) = \lim_{n}(Tx_{n},z)=(y,z),\;\;\; z \in \mathcal{H}. $$ Because $(Tx-y,z)=0$ for all $z\in\mathcal{H}$, then $Tx=y$. So the graph of $T$ is closed. Similarly the graph of $S$ is closed.

If you know the uniform boundedness principle (Banach-Steinhaus theorem), you could first show that the set $C=\{Tx:\|x\|\le 1\}$ is weakly bounded.
Then, if you apply Banach-Steinhaus theorem properly (in a subtle way), you get that $C$ is bounded (which tells you that $T$ is continuous).

Here's a proof outline that directly uses the uniform boundedness principle rather than the closed graph theorem. This is probably more or less equivalent to Mizar's idea. Consider the family of linear functionals $$ T_x(y) = \frac{1}{\Vert x \Vert}(Ax,y), $$ indexed by $x\in H$. Define $S_x$ analogously, except with the $S$ on the other side of the inner product. Try to find the norms of $T_x$ and $S_x$; you'll see why they are interesting for this problem. Moreover these operators are related by a simple inequality due to $T$ and $S$ being adjoint. Find this inequality, and you'll find yourself able to use the uniform boundedness principle.