Where does this proof use the fact that $f'$ is continuous? I'm reading part of a proof where $\mathcal{O}$ be an open subset of $\mathbb{R}$, and $f$ is continuously differentiable. 
Then the proof starts as follows:

Suppose $f'(x_0) > 0$. Since $x_0$ is an interior point of $\mathcal{O}$ and the function $f' : \mathcal{O} \rightarrow \mathbb{R}$ is continuous, we can choose $r > 0$ such that the closed interval $[x_0 - r, x_0 + r] \subseteq \mathcal{O}$

I don't understand how this follows from continuity. Since $x_0 \in \text{ int } \mathcal{O}$, we can choose an open ball about $x_0$ contained in $\mathcal{O}$. But how can we include the endpoints? I'm guessing that this is where continuity of $f'$ comes into play, but I'm not sure.
 A: Since $x_0 \in \mathcal{O}$ where the latter is open, there exists an open interval $(x_0-\delta,x_0 + \delta) \subseteq \mathcal{O}$. In particular,
$$
\left[x_0-\frac{\delta}{2},x_0+\frac{\delta}{2}\right] \subset (x_0-\delta,x_0 + \delta) \subseteq \mathcal{O}.
$$
You are also correct in saying that continuity is not required for this. Although as pointed out in the comments, the continuity of $f^\prime$ is used elsewhere in the proof you are referring to. Namely, we use the continuity of $f^\prime$ together with the fact that $f^\prime(x_0) > 0$ to ensure that $f^\prime > 0$ on $\left[x_0-\frac{\delta}{2},x_0+\frac{\delta}{2}\right]$.
To see why this is possible, it will be enough to establish the following result:

Lemma. Let $O\subseteq \mathbb{R}$ be an open set and $g :O \to \mathbb{R}$ a continuous function on $O$. Assume further that $x_0 \in O$ is such that $g(x_0) > 0$. Then, there exists a compact interval 
  $$
\left[x_0-{\delta},x_0+{\delta}\right] \subseteq {O}
$$
  on which $g$ is strictly positive.

To prove this, let $\epsilon := g(x_0)/2 > 0$. Using the continuity of $g$, we can find $\delta_1 > 0$ such that
$$
g(x_0) - g(x) \leq |g(x)-g(x_0)| < \epsilon = \frac{g(x_0)}{2}
$$
for all $x \in {O}$ with $|x-x_0| < \delta_1$. Hence, one has
$$
g(x) > \frac{g(x_0)}{2} > 0
$$
for all these $x$. On the other hand, because $O$ is open, we can find $\delta_2 > 0$ such that 
$$
\left[x_0-{\delta_2},x_0+{\delta_2}\right] \subseteq O.
$$
Taking $0 < \delta < \min(\delta_1,\delta_2)$ then proves the lemma.
