Let $S$ be the closed unit ball of $B(H)$, the bounded operators on the Hilbert space $H$. I want to show that the relative strong topology on $S$ is metrizable.
Attempt: I have already established that the strong topology on $S$ is separable, so we may select a norm-dense sequence $(x_n)_n$ in $S$. Define a metric $d$ on $S$ by $$d(u,v) = \sum_{n=1}^\infty \frac{\Vert(u-v)x_n\Vert}{2^n}$$ This series converges since $u,v \in S$ and it is easily verified that this is in fact a metric.
Denote the metric topology on $S$ by $\tau_d$ and the strong topology on $S$ by $\tau_s$. We want to show $\tau_d = \tau_s $.
If $u_\lambda \to u$ in $\tau_d$, then it is easily seen that $\Vert (u-u_\lambda)x_n \Vert \to 0$ when $\lambda \to \infty$, hence by density for all $x \in H$ we have $\Vert (u-u_\lambda)x\Vert \to 0$ which means that $u_\lambda \to u$ in $\tau_{s}$. It follows that $\tau_s \subseteq \tau_d$.
If $u_\lambda \to u$ in $\tau_s$, then I want to show that
$$\sum_{n=1}^\infty \frac{\Vert (u-u_\lambda)x_n\Vert}{2^n} \to 0$$
This boils down to interchanging the infinite sum and limit of the net. This makes me thing to use dominated convergence, which will work if we can show that we can replace nets by sequences. So, if we can prove the strong topology on $S$ (or more generally $B(H)$) is first countable, then I will be done. For this, it suffices to find a strong neighborhood basis of the origin
How can I proceed?
Your $x_n$ should be in $H$, I think.
That said, you can estimate the limit directly. Fix $\varepsilon>0$, and $m$ such that $\sum_{n>m}2^{-n}<\varepsilon/4$. Choose $\lambda_0$ such that, for all $\lambda>\lambda_0$, we have $\|(u-u_\lambda)x_n\|<\varepsilon/2$, $n=1,\ldots,m$. Then \begin{align} \sum_n\frac{\|(u-u_\lambda)x_n\|}{2^n} &\leq\sum_{n=1}^m\frac{\|(u-u_\lambda)x_n\|}{2^n}+\sum_{n>m}\frac{2}{2^n}\\[0.3cm] &\leq \frac\varepsilon2\,\sum_{n=1}^m2^{-n}+\frac\varepsilon2\\[0.3cm] &<\varepsilon. \end{align}