Let $X$ be a dual Banach space, i.e. $X=(X_*)^*$ for some Banach space $X_*$. Consider the weak* topology of $B(X)$, i.e. the topology of pointwise convergence on $X$ endowed with the $\sigma(X,X_*)$-topology.
Consider the set $B_{w^*}(X)$ of $w^*$-continuous bounded operators on $X$. Is it a closed subset of $B(X)$ for the weak* topology of $B(X)$ ?
By this answer, you are looking at the set $$ B_{w^{\ast}}(X) = \{T^{\ast} : T \in B(X_{\ast})\} $$ which is convex.
Now if $\sigma$ denotes the weak-$\ast$-topology on $B(X)$, then $(B(X),\sigma)$ is a topological vector space whose topology is generated by the family of semi-norms given by $$ p_{f,y}(T) := |T(f)(y)| $$ for $f\in X, y\in X_{\ast}$. Hence, by a consequence of the Hahn-Banach separation theorem (see this answer), $B_{w^{\ast}}(X)$ is closed with respect to $\sigma$ iff it is closed in the norm topology of $B(X)$.
But this last fact is easy to prove:
Proof: If $\{T_n\} \subset B(X_{\ast})$ and $S\in B(X)$ such that $\|T_n^{\ast} - S\|_{B(X)} \to 0$, then we wish to show that $S$ is weak-$\ast$-continuous. To see this, choose a net $f_{\alpha} \in X$ such that $f_{\alpha}(y) \to f(y)$ for all $y\in X_{\ast}$. By the principle of uniform boundedness, $\exists M > 0$ such that $$ \sup_{\alpha} \|f_{\alpha}\|, \|f\| < M $$ (Here we need the fact that $X_{\ast}$ is a Banach space). Then for any $\epsilon > 0$ and fixed $y\in X_{\ast}$, choose $N \in \mathbb{N}$ such that $$ \|T_N^{\ast} - S\|_{B(X)} < \frac{\epsilon}{3M\|y\|} $$ and $\beta$ such that $$ \|f_{\alpha}(T_N(y)) - f(T_N(y))| < \epsilon/3 $$ for all $\alpha \geq \beta$. Then, for all $\alpha\geq \beta$, we have \begin{equation*} \begin{split} |S(f_{\alpha})(y) - S(f)(y)| &\leq |S(f_{\alpha}(y)) - T_N^{\ast}(f_{\alpha})(y)| + |T_N^{\ast}(f_{\alpha})(y) - T_N^{\ast}(f)(y)| + |T_N^{\ast}(f)(y) - S(f)(y)| \\ &< \epsilon \end{split} \end{equation*} and so $S(f_{\alpha}) \to S(f)$ pointwise as required.