Is this proof regarding boundary and closed sets correct?

Is this proof regarding boundary and closed sets correct?

I've tried to prove the following: Let $A\subset \mathbb{R}^n$, then $A$
is closed if and only if $\partial A\subset A$. My proof is as follows:
Notice first that we always have $\partial A \subset
\operatorname{Cl}(A)$, because $p \in \partial A$ implies that every ball
$B(p;r)$ contains points of $A$ and of $\mathbb{R}^n\setminus A$, hence it
contains non-empty intersection with $A$ and hence $p \in
\operatorname{Cl}(A)$. Now, if $A$ is closed, $A = \operatorname{Cl}(A)$
and hence $\partial A \subset A$.
Suppose now that $\partial A \subset A$, since we want to show that $A$ is
closed, we are going to show that $\mathbb{R}^n \setminus A$ is open.
Consider $p \in \mathbb{R}^n \setminus A$, then $ p \notin A$ and hence by
hypothesis $p \notin \partial A$. So, since a point is in $\partial A$
implies that every open ball $B(p;r)$ contains points of $A$ and of
$\mathbb{R}^n\setminus A$ it must be the case that there is a ball
$B(p;r)$ without points of $A$ or without points of $\mathbb{R}^n
\setminus A$. Since $p \in \mathbb{R}^n \setminus A$ and the ball is
centered at $p$, the second possibility is always impossible, every ball
will have at least the point $p \in \mathbb{R}^n \setminus A$, so there
must be the cases that exists an open ball $B(p;r)$ without points of $A$,
and thus $B(p;r)\subset \mathbb{R}^n\setminus A$ showing that $A$ is
closed since the complementary is open.
Since the proof was too simple I felt that it could be wrong. Is this
proof ok? Is there something wrong with it? Thanks very much in advance!