Let $X$ be a locally convex Hausdorff space. Then $X$ injects into $X^{**}$ via the canonical map $\kappa: X\to X^{**}$. Now, $X^{**}$ carries the weak* topology. Let $B$ be a bounded set in $X$. Is $\kappa[B]$ weak* pre-compact in $X^{**}$?
By the Banach-Alaoglu theorem, this is the case when $X$ is a normed space.
Beyond Banach spaces, it is not obvious which topology you use on the dual $X^*$ to determine the second dual. The standard choice is the strong topology $\beta(X^*,X)$ of uniform convergence on all bounded subsets $B$ of $X$ (where, of course, boundedness of $B$ in $X$ means bounded with respect to each continuous semi-norm). The semi-norms of $\beta(X^*,X)$ are $p_B(\phi)=\sup\{|\phi(x)|:x\in B\}$ (which are real-valued because of the continuity of $\phi$) and their unit balls are the (absolute) polars $B^\circ$. These are thus $0$-neighbourhoods in $(X^*,\beta(X^*,X))$ so that their polars $B^{\circ\circ}$ in $(X^{**},\sigma(X^{**},X^*))$ are compact by Banach-Alaoglu. It remains to check that $\kappa[B]$ is contained in $B^{\circ\circ}$ so that it is even relatively weak$^*$-compact and hence also weak$^*$ pre-compact.
