Let $(a,b) \in \mathscr H \times \mathscr H$ be fixed.
So we have to prove that for a given $\epsilon \gt 0$, we can find $\delta_1 \gt 0$ and $\delta_2 \gt 0$ such that $\lvert \langle x,y\rangle - \langle a,b\rangle\rvert \lt \epsilon$ whenever $\lVert x-a\rVert \lt \delta_1$ and $\lVert y-b\rVert\lt \delta_2$ and to do this we have to use the Cauchy-Schwarz inequality $\lvert \langle x,y\rangle\rvert \le \lVert x\rVert \lVert y\rVert$.
(Note that "$\langle\,,\rangle$" denotes the inner product and "$\lVert \cdot\rVert$" denotes the norm induced by it.)
Let $\epsilon \gt 0$ be given.
Then, I start with
\begin{align} \lvert \langle x,y\rangle - \langle a,b\rangle\rvert &= \lvert\langle x,y\rangle - \langle a,b\rangle + \langle x,b\rangle - \langle x,b\rangle\rvert = \lvert\langle x,y-b\rangle + \langle x-a,b\rangle\rvert\\ &\le \lvert\langle x,y-b\rangle\rvert + \lvert\langle x-a,b\rangle\rvert \le \lVert x\rVert \lVert y-b\rVert + \lVert x-a\rVert \lVert b\rVert. \end{align}
Now I choose $\delta_1=\frac {\epsilon}{\lVert b\rVert}$ and $\delta_2=\frac {\epsilon}{\lVert a\rVert}$.
Hence, $\lVert x\rVert - \lVert a\rVert \le \lVert x-a\rVert \lt \delta_1$ and $\lVert y-b\rVert \lt \delta_2$
$\Rightarrow$
$\lvert\langle x,y\rangle - \langle a,b\rangle\rvert \le (\lVert a\rVert + \delta_1)\delta_2 + \delta_1 \lVert b\rVert = \bigl(\lVert a\rVert + \frac {\epsilon}{\lVert b\rVert}\bigr)\frac {\epsilon}{\lVert a\rVert} + \epsilon = 2\epsilon + \frac {\epsilon^2}{\lVert a\rVert \lVert b\rVert}$.
So we conclude that $\langle\, ,\,\rangle$ is continuous.
Am I on the right track so far? Is there a more simpler way to do this (Using Cauchy-Schwarz inequality obviously.)?