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

Set $S$ (on a metric space) is open if $\forall x \in S$, $\exists
 \delta > 0,$ s.t. $\thinspace \overline B_\delta(x) \subset X$

I am a little bit thrown off by the closed ball instead of open ball definition of open set. Can someone verify the above statement and show how it is same as open ball.
 A: 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 $0<d<\delta$, $B_d(x)\subset S$.
A: 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.
A: Both other answers are correct, but they are not getting to the metric content of the problem...
Let $X$ be a metric space, $S \subset X$, and $x \in S$.  If $S$ is open, then $X \setminus S$ is closed, so $\mathrm{dist}(x, X \setminus S)$ exists and is positive.  Otherwise, if $x \in \partial S$, $\mathrm{dist}(x, X \setminus S)$ is only nonnegative.  Set $\delta= \frac{1}{2} \mathrm{dist}(x, X \setminus S)$.  Then $\overline{B_\delta (x)} \subset S$.
Note that $\delta = \frac{1}{2}\mathrm{dist}(\dots)$ is not essential.  We can choose and $\eta \in (0,1)$ and $\delta = \eta \mathrm{dist}(\dots)$ works as well.  In any event, if $x$ may be chosen on $\partial S$, $\delta$ is forced to $0$ and the hypothesis ("$\forall x \in S, \exists \delta > 0 \dots$") does not hold.
A: 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.
