It is known that the unit ball in $B(X)$ where $X$ is a reflexive space is compact in the weak operator topology (see e.g Compactness of unit ball in WOT of B(X))
Is it true that the unit ball in $B(X)$ where $X$ is a separable reflexive space is sequentially compact in the weak operator topology?
Yes: the weak operator topology on the unit ball $B$ of $B(X)$ is metrizable, so compactness implies sequential compactness. To prove this, note that by definition the weak operator topology is the coarsest topology that makes the functional $\mu_{x,y}:T\mapsto \langle T(x),y\rangle$ continuous for each $x\in X$ and $y\in X^*$, where $\langle\cdot,\cdot\rangle$ is the duality pairing between $X$ and $X^*$. Note though that if $T\in B$ then $$|\mu_{x,y}(T)-\mu_{x',y'}(T)|\leq |\langle T(x-x'),y\rangle|+|\langle T(x'),y-y'\rangle|\leq \|x-x'\|\|y\|+\|x'\|\|y-y'\|$$ since $\|T\|\leq 1$.
Now let $D$ be a countable dense subset of $X$ and $E$ be a countable dense subset of $X^*$. I claim that the weak operator topology on $B$ is actually the same as the coarsest topology that makes the functionals $\mu_{x,y}$ for $x\in D$ and $y\in E$. Indeed, by the inequality above, for any $x'\in X,y'\in X^*$, we can approximate $\mu_{x',y'}$ uniformly on $B$ by functionals of the form $\mu_{x,y}$ for $x\in D,y\in E$, and so for any basic open neighborhood in the weak operator topology we can find a smaller neighborhood defined by such functionals $\mu_{x,y}$. Since there are only countably many such functionals $\mu_{x,y}$, and they are all bounded on $B$, the topology they generate on $B$ is metrizable.
