If $f$ is increasing and continuous at $x$, then there exists two sequences$\{ a_n \}$ and $\{ b_n \}$, $a_n < x < b_n$

The following question is extracted from Royden's Real Anlysis $4$th edition, question $36$ at page $53$:

Let $f$ be an increasing function on the open interval $I$. For $x_0 \in I,$ show that $f$ is continuous at $x_0$ if and only if there are sequences $\{ a_n \}$ and $\{ b_n \}$ in $I$ such that for all $n$, $a_n < x_0 < b_n$ and $\lim_{n \rightarrow \infty}{[f(b_n) - f(a_n)]} = 0.$

I am trying to prove the $(\Rightarrow)$ direction.

I have been trying to do the following:

Since $f$ is continuous at $x_0$, for any $n \in \mathbb{N}$, there exists $\delta_n >0$ such that for all $x$, $|x - x_0| < \delta_n \Rightarrow |f(x) - f(x_0)| < \frac{1}{n}.$

However, I fail to see that why must there $a_n < x_0 < b_n .$ In the first place, how do we know that $a_n$ and $b_n$ are different from $x_0$?

Because $$I$$ is open, there is an $$r > 0$$ for which the interval $$(x_0 - r,x_0 + r)$$ is contained in $$I$$. Choose a natural number $$N$$ such that $$1/N < r$$. Then the sequences defined by $$a_n = x_0 - 1/(N + n)$$, and $$b_n = x_0 + 1/(N + n)$$ are in $$I$$ such that for each $$n$$, $$a_n < x_0 < b_n$$. Moreover, $$\{a_n\}$$ and $$\{b_n\}$$ converge to $$x_0$$ so that by Proposition 21 of Section 1.6, $$\lim_{n\to\infty} f(a_n) = f(x_0) = \lim_{n\to\infty} f(b_n)$$. Thus, by Theorem 18 of Section 1.5, $$\lim_{n\to\infty} (f(b_n) - f(a_n)) = f(x_0) - f(x_0) = 0$$.
Hint: Start with $$a_n = x_0 - \frac{1}{n}$$ and $$b_n = x_0 + \frac{1}{n}$$. Then use the fact that $$I$$ is open to show that at some point both sequences are in $$I$$. Conclude using continuity at $$x_0$$.