The interior of a set is open and the boundary and closure of a set is closed

So I am pretty sure my proof is correct, just want to verify for correctness and rigor.

Let $\displaystyle a\in int(S)$. Then there exists $\displaystyle \epsilon > 0$ such that $\displaystyle B(\epsilon , a) \subset S$. Now suppose $\displaystyle x\in B(\epsilon , a)$. Since $\displaystyle B(\epsilon, a)$ is open, there exists $\displaystyle \delta > 0$ such that $\displaystyle B(\delta , x) \subset B(\epsilon , a) \subset S$; in other words, for any $\displaystyle x\in B(\epsilon , a)$ there is a ball centered at $\displaystyle x$ contained solely in $\displaystyle S$, which means $\displaystyle x\in int(S)$. So given $\displaystyle a\in int(S)$ there exists $\displaystyle r > 0$, namely $\displaystyle r = \epsilon$, such that $\displaystyle B(r,a) \subset int(S)$, so $\displaystyle int(S)$ is open.

So now consider $\displaystyle \partial S$. Then $\displaystyle (\partial S)^{c} = int(S) \cup int(S^{c})$ since if $\displaystyle x$ is not a point such that every ball centered at $\displaystyle x$ intersects $\displaystyle S$ and $\displaystyle S^{c}$, then there exists some ball centered at $\displaystyle x$ that is either contained solely in $\displaystyle S$ or $\displaystyle S^{c}$. So we have that $\displaystyle (\partial S)^{c}$ is the union of two open sets, which is open, so $\displaystyle \partial S$ is closed. Similarly, if we consider $\displaystyle cl(S)$, then $\displaystyle (cl(S))^{c} = S^{c}\cap(int(S)\cup int(S^{c})) = int(S^{c})$, which is open, so $\displaystyle cl(S)$ is closed.

Q.E.D.

So I think this is a perfectly valid and rigorous proof, but any feedback would be appreciated (I am somewhat unsure on the closed portions of the proof, especially the claims about what the complements of the boundary of S and the closure of S actually equal). Thanks.