Prove that a metric space $S$ is disconnected iff $\exists$ a non empty subset $A$ of $S$ s.t. $A \neq S$ and $A$ is both open and closed.
Below is my proof. Any input would be appreciated. I'm not totally convinced I have the correct notation for the "if" portion ...
$\Leftarrow$ if: $\qquad \quad$ Assume $S$ is disconnected and $U$ and $V$ form $A \subset S$.
$\qquad \quad \quad \quad$ Since $U \cap V \neq \emptyset$ and $U \cup V = A$, we know $V = A \setminus U$. And since $U$ is open, $V$ is $\qquad \quad \quad \quad$ closed.
$\qquad \quad \quad \quad$ Since $U \neq \emptyset$, $V \neq A \Rightarrow A \neq S$.
$\qquad \quad \quad \quad$ Thus, $A$ is a nonempty subset of $S$ that is both open and closed.
$\Rightarrow$ only if: $\quad \ $ Assume $A \neq S$ and $A$ is nonempty, open and closed.
$\qquad \quad \quad \quad$ Since $A$ is closed, $S\setminus A$ is open.
$\qquad \quad \quad \quad$ Since $ A < S , S \setminus A \neq \emptyset$
$\qquad \quad \quad \quad$ So, $(S \setminus A) \cup A = S$ and $(S \setminus A) \cup A = S$
$\qquad \quad \quad \quad$ Thus, $S$ is disconnected