The Skorohod's representation theorem says:
Let $X, X_1, X_2,...$ be S-valued random elements, where (S, p) is a separable metric space, such that $X_n \xrightarrow{d} X$. Then there exists a common probability space with some random elements $Y, Y_1, Y_2,...$ defined on it such that $Y_n = X_n$ in distribution $\forall n$ and $Y = X$ in distribution, and $Y_n$ converges almost surely to $Y$.
Every proof I found involves the process of the following construction: for all $x \in S$, for any $\epsilon>0$, one can always pick $r$ with $0<r<\epsilon$ such that $P(X\in \partial B(x,r))=0$.
My question might be stupid: why is it always possible to pick such an open ball such that its boundary has measure $0$ in $P_X$? (in other words, an $X$-continuity set, which is defined as: a set $A\in B(S)$ such that $P(X\in \partial A)=0$)
For complete proof, see resource: Kallenberg's Foundations of Modern Probability (2002) p79; or Billingsley's Convergence of Probability measures (1999) p70.
$\partial B(x,r)\subset \{y: d(y,x)=r\}$. It is obvious from this that $\partial B(x,r), 0<r<\epsilon$ is a disjoint collection of measurable sets. In a probability space you can only have countably many disjoint sets with positive measure. [If you have not seen this before I will add a proof]. If $P(\partial B(x,r))>0$ whenever $0<r<\epsilon$ you will uncountably many sets with positive measure, a contradiction. Hence, $P(\partial B(x,r))=0$ for some $r \in (0,\epsilon)$.
Proof of $\partial B(x,r)\subset \{y: d(y,x)=r\}$: Let $y \in \partial B(x,r)$. By definition of boundary there exist sequence $(y_n)$ and $(z_n)$ such that $d(y_n,x) <r, d(z_n,x) \geq r$ and $y_n \to y, z_n \to y$. By continuity of the distance function we get $d(y,x) \leq r $ and $d(y,x) \geq r$ so we must have $d(y,x)=r$.