proof for a continuous distribution function where a single point has 0 probability of occurring

Here is a proof using the continuity property of probabilities(property which I understand) that for a continuous distribution function, a single point has 0 probability of occurring.
I understand fully this proof until the last line where i'm not completely sure why this follows. Is it because $F_X$ $(a-1/n)$ converges to a as n tends to infinity and hence the expression is equal to 0? (how is $F_x$ being continuous connected to this deduction?)

Answer

Your guess about the reason is correct. Remember that a function $f : \mathbb{R} \to \mathbb{R}$ is continuous at a point $a$ if and only if, for every sequence $(x_n)_{n=1}^{\infty}$ such that $\lim_{n\to\infty}x_n = a$, we have $\lim_{n\to\infty}f(x_n)=f(a)$. In particular, if $F_X$ is continuous, then since $\lim_{n\to\infty}(a- 1/n) = a$, we get $\lim_{n\to\infty}F_X(a-1/n) = F_X(a)$.