EDIT: I rewrite my answer, since it appears it was not clear enough. I will try to go in more details even though I think this is not really interesting.
We define $F(x) := P(X \leq x)$ and we want to prove $F(x-) = P(X < x)$.
Consider a sequence $(a_n)$ such that $\lim_{n \rightarrow \infty} a_n = x$, $a_n < x$ for all $n$. For the sake of simplicity, let's admit $a_n$ is increasing ($a_{n+1} \geq a_n$).
Now, let's consider the sequence $(b_n)$ given by $b_n := F(a_n)$. By definition of $F$, we have $b_n = P(X \leq a_n)$. Thus $(b_n)$ is a sequence in [0,1]. Since $b_{n+1} = P(X \leq a_{n+1}) \geq P(X \leq a_n) = b_n$, we have that the sequence $b_n$ is also increasing. Hence, it converges. Now let's calculate the limit (let's call it $b$).
We know that $P(X \leq a_n) \leq P(X < x)$, so $b_n \leq P(X < x)$ for all $n$. Thus this should remain true for the limit $b$. You might argue that you are not sure why this should remain true for the limit and why we could not have $b = P(X \leq x)$. Here comes the $\varepsilon$ argument. If $P(X \leq x) = P(X < x)$ then there is no problem, so suppose $P(X \leq x) \ne P(X < x)$. In that case, let $\varepsilon := \frac{1}{2}\left(P(X \leq x) - P(X < x)\right)$.
By definition of the limit, you should find infinite members of the sequence in $[b- \varepsilon, b + \varepsilon]$. But this can't be, because all the $b_n$ are smaller than $P(X < x)$ and thus at distance at least 2 $\varepsilon$ from the limit. This proves that if $P(X \leq x) \ne P(X < x)$, then $b \ne P(X \leq x)$.
Ok, so we know $b \leq P(X < x)$. We still need to prove there is an equality here. I will not go into further technical details here, the intuition should tell you that since $a_n \rightarrow x$, you can come as near to $P(X < x)$ as you want with the sequence $b_n$ (consider $\varepsilon = \frac{P(X < x) -b}{2}$ and see what this implies bla-bla-bla).
In the beginning I said we could consider $a_n$ to be increasing. Suppose it is not. Then you can construct a new sequence $c_n$, such that $\lim c_n = x$, $c_n \leq a_n$ for all $n$ and $c_n$ is increasing. Then you compare the sequence $d_n := F(c_n)$ with the sequence $b_n$ and bla-bla-bla. This really becomes technical, but it works. So $F(x-) = P(X < x)$.
by the way, sorry if I can't write my sequences properly, I can't get { } in math mode, even with 1 backslash, 2 backslaches or 3 backslaches before the {. This really annoys me.