Are bounded sets always weakly metrizable in reflexive separable spaces? It is known that if a Banach space is reflexive and separable, its unit ball is weakly metrizable. 
My question is about the generalization of this property : 
1) Is it true that for all reflexive separable locally convex space, bounded sets are weakly metrizable? 
2) If that's true, is there a way to explicitly construct a distance for the weak-topology on any bounded set? 
 A: 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.
A: No. Let $X$ be any separable Banach space s.t. $X^*$ is non separable and consider the space $X^{**}$ with  the weak$^*$ topology.
Edit: As Jochen and vitava point out in the comments below, this only gives   a semi-reflexive example even if you use the Mackey topology on $X^{**}$ with respect to the duality $(X^{**}, X^*)$.  It is not reflexive under the classical definition of reflexive for LCTVS.
