No. Let $I$ be an index set with the cardinality of the continuum. Endow $X=\mathbb R^I$ with the product topology. According to (a particular case of) the Hewitt-Marczewski-Pondiczery theorem (which is 2.3.15 in Engelking's *General Topology*) $X$ is separable. Moreover, it is semi-reflexive (by Tychonov) and barrelled (because barrelledness is stable w.r.t. products, proposition 4.2.5 in *Barrelled Locally Convex Spaces* of Bonet and Perez Carreras). Therefore, $X$ is reflexive. The set $B=[-1,1]^I$ is bounded. Assume that it is weakly metrizable (by the way, $X$ carries its weak topology). Then it has a countable basis of $0$-neighbourhoods and since every $0$-neighbourhood in $X$ contains one of the form $\lbrace x\in X: |\varphi_i(x)|<\varepsilon, i=1,\ldots,n \rbrace$ for some $n\in \mathbb N$, $\varphi_1,\ldots,\varphi_n \in X'$, and $\varepsilon>0$, we find a sequence of $\phi_n\in X'$ such that
$$
B\cap \bigcap_{n\in\mathbb N} \text{kern}(\phi_n) = \lbrace 0 \rbrace.
$$
But each $\phi_n$ only depends on finitely many coordinates, i.e., it is of the form $\phi_n((x_i)_{i\in I})= \sum_{i\in E_n} a_{n,i} x_i$ for some *finite* $E_n \subseteq I$, and it is enough to consider $j\in I\setminus \bigcup_{n\in\mathbb N} E_n$ and $b=(\delta_{i,j})_{i\in I} \in B$ to get a contradiction.

dualis separable to have the bounded sets weakly metrizable. If your space $X$ is metrizable separability of $(X',\sigma(X',X))$ follows from the separability of $X$, but in general, I doubt. As a reference I recommend the bookBarrelled Locally Convex Spacesof Bonet and Perez-Carreras, in particular, section 2.5. $\endgroup$