We say that a sequence of random variables $X_1,X_2,\cdots$ converges in law to a random variable $X$ if $F_{X_n}(t) \to F_X(t)$ (where F is the cdf) for every $t$ that is a continuity point of $F_X$, i.e. that is such that $\Bbb P(X = t) = 0$.
An equivalent criteria I was given is : let $X_1, X_2, . . .$ be a sequence of random variables. They converge to a random variable $X$ in law if and only if for every $a < b$ with $\Bbb P(X = a) = \Bbb P(X = b) = 0$ we have that $\Bbb P(X_n \in (a, b)) \to \Bbb P(X \in (a, b))$
The proof of the $\Longrightarrow$ implication I was given is :
If $(X_n)_{n≥1}$ converge in law to $X$ then by definition $F_{X_n}(t) \to F_X(t)$ for any continuity point $t$ of $F_X(t)$. In particular, if $\Bbb P(X = a) = \Bbb P(X = b) = 0$, then the points $a, b$ are such continuity points. We can write $$\Bbb P(X \in (a, b)) = F_X(b) − F_X(a) = \lim_{n\to \infty}(F_{X_n}(b) − F_{X_n}(a)).$$ But now $\Bbb P(X_n \in (a, b)) = F_{X_n}(b^−)−F_{X_n}(a)$. It suffices to now see that $ \lim_{n\to \infty}F_{X_n}(b^-) = \lim_{n\to \infty}F_{X_n}(b)$. But this follows from the fact that $b$ is a continuity point as for every $\epsilon > 0$ we have that $$F_{X_n}(b−\epsilon) ≤ F_{X_n}(b^−) ≤ F_{X_n}(b)$$ and if $b − \epsilon$ is also a continuity point, we deduce $$F_X(b − \epsilon) ≤ \liminf_{n\to \infty}F_{X_n}(b^−) ≤ \limsup_{n\to \infty}F_{X_n}(b^−) ≤ F_X(b),$$ which letting $\epsilon \to 0$ gives the desired equality.
I do not understand the proof starting from the sentence in bold, there is no assumption telling us that $b$ is a continuity for $F_{X_n}$, also $b - \epsilon$ is also another assumption they make that is not in the initial statement. Can someone explain what is being done here ?