Is this statement true: a set is open if every point has a closed ball contained inside of the set

Hint: show that $\overline{B}_r(x) \subseteq B_\delta(x)$ for every $r < \delta$.

And clearly $B_\delta(x) \subseteq \overline{B}_\delta(x)$ for all $x$ and all $\delta$.

So every closed ball contains a smaller open ball with the same centre, and every open ball contains a smaller closed ball with the same centre as well.

So for openness we can both use open balls and closed balls.

Pick $x\in S$. Then there's a closed ball around it that's contained in S. There's an open ball with a slightly smaller radius that's centered around $x$, so it's also contained in S. So $x$ is an interior point of $S$.

More formally, for $x\in S$, there exists some $\delta>0$ so that $\overline{B_\delta(x)} \subset D$. Then for $0<d<\delta$, $B_d(x)\subset S$.

Observe that you can write the set

$$S = \bigcup_{x \in S} B_{\delta/2}(x) $$

as a union of open sets, which is therefore open, as desired. QED.