In a topological space, is a perfect set (i.e. closed with no isolated points) always the boundary of some set?
A topological space is irresolvable if there is no dense subset whose complement is also dense. If $X$ is irresolvable and dense-in-itself, then $X$ is a perfect subset of $X$ which is not the boundary of any subset of $X$. Irresolvable dense-in-itself spaces have been constructed by Hewitt and others:
E. Hewitt, A problem of set-theoretic topology, Duke Math. J. 10 (1943), 309-333.
K. Padmavally, An example of a connected irresolvable Hausdorff space, Duke Math. J. 20 (1953), 513-520.
Douglas R. Anderson, On connected irresolvable Hausdorff spaces, Proc. Amer. Math. Soc. 16 (1965), 463-466.
For an easy example of an irresolvable T$_1$-space with no isolated points, take an infinite set $X$ and a nonprincipal ultrafilter $\mathcal U$ on $X$, and topologize $X$ by making $\mathcal U$ the collection of all nonempty open sets. This is a door space; clearly, a nonempty door space is irresolvable.