Is the claim in the title true? If so, can someone link or provide a proof? Also, out of curiosity, if the claim is true for the real line, is it true for $\mathbb{R}^n$, or some more general spaces?
For example, Consider the subset $[0,2]\subset \mathbb{R}$, and a cdf $F$ on this subset. We know the limit from the right of $F$ exists and equals $F(x)$ (we approaching $x$). So we just have to show that if $F$ has no atoms then the limit from the left approaching $x$ is $F(x)$.
This seems trivial but I can't think of how to formally write it. (Intuitively, the function can't suddenly "jump up" at the limit.
Let $X$ be a random variable with cdf $F$. Every cdf is right continuous at every point $x$ since $(X\leq x+n^{-1})\downarrow (X\leq x)$ so $$ F(x)=P(X\le x)=\lim_{n\to\infty}P(X\le x+n^{-1})=\lim_{y\downarrow x}F(y) $$ by measure continuity from above. Similarly, $(X\leq x-1/n)\uparrow (X\lt x)$ whence $$ P(X<x)=\lim_{n\to\infty}P(X\le x-n^{-1})=\lim_{y\uparrow x}F(y)=F(x-)$$ by measure continuity from below. Now observe that $$ P(X=x)=P(X\leq x)-P(X<x)=F(x)-F(x-) $$ Thus $P(X=x)=0$ iff $F(x+)=F(x)=F(x-)$ i.e. iff $F$ is continuous at $x$.