Are Borel subsets of Polish spaces Polish? Are Borel subsets of Polish spaces Polish? I would imagine not, since I would think that would be noted in the Wikipedia page on them, and it isn't.
This question arises in connection to trying to show that measures of the sort asked about in this question don't exist besides the trivial one.
 A: A simple counterexample is the set of rationals $\mathbb{Q}$ as a subset of $\mathbb{R}$.  It has no isolated points, therefore every point of $\mathbb{Q}$ is a nowhere dense subset of $\mathbb{Q}$.  But $\mathbb{Q}$ is the countable union of such points, which if it were Polish would contradict the Baire category theorem.
A: Only $G_\delta$ subsets of Polish spaces are Polish (equipped with the subspace topology). Nevertheless, if you are given a Polish space $X$ and its Borel subset $B$ (not necessarily $G_\delta$) then you can find a Polish topology on $X$ giving the same Borel $\sigma$-algebra as the original topology on $X$, in which $B$ is clopen (hence Polish if considered with topology induced by the new topology on $X$).
A: No. A subset of Polish space is Polish in the relative topology if and only if it is $G_\delta$. The inverse implication is attributed here to Alexandroff. For a proof, check Kechris' book, Theorem 3.11 in section 3.C.
In an uncountable Polish space without isolated points, any countable dense set can't be Polish, by the Baire Category Theorem (and the previous theorem): if it were $G_\delta$, it is the intersection of a countable family of dense open sets, and hence it can't be (included) in the countable union of closed sets with empty interior.
