Helly's selection theorem for distribution functions states that for every sequence $F_n$ of distribuion functions, there is a subsequence $F_{n(k)}$ and a right continuous nondecreasing function $F$ so that $\lim_{k \to \infty} F_{n(k)} (y) \to F(y)$ at all continuity points of $F$.
I am wondering why this implies that $0 \leq F(y) \leq 1$ for all $y$. For any finite $y$ such that $y$ is a continuity point of $F$, I do see why that is true, because in that case $\lim_{k \to \infty} F_{n(k)} (y) \to F(y)$ by the theorem and $0 \leq F_{n(k)} (y) \leq 1$ for all $k$. However, what if $y$ is not a continuous point of $F$ or $y$ is $\infty$ or $-\infty$? I do feel that we can use the fact that $F$ is right continuous and nondecreasing and the fact that any monotone function that takes values in $\mathbb R$ has at most a countable number of discontinuities, implying that the continuity points of $F$ is dense in $\mathbb R$.
As you say $F$ is nondecreasing and has at most a countable number of discontinuities. So for all $y\in\mathbb R$, there exist $x,z\in\mathbb R$ such that $x<y<z$ and $x$ and $z$ are continuity points of $F$. Since $x$ and $z$ are continuity points of $F$, $0\le F(x)$ and $F(z)\le1$. And by monotonicity of $F$, $F(x)\le F(y)$ and $F(y)\le F(z)$. Hence $0\le F(y)\le 1$.
Are you sure your statement allows $y=+\infty$ or $y=-\infty$? If that's the case, then they are continuity points by definition, because by definition you would define $F(+\infty)=\lim_{x\to+\infty}F(x)$, and similarly for $-\infty$. Since $F_n(+\infty)=1$ and $F_n(-\infty)=0$, you deduce that $F(+\infty)=1$ and $F(-\infty)=0$.