# Proof :

Let (X, d) be a metric space and suppose that it is disconnected.
Claim: $\exists$ a non-empty proper subset of X which is both open & closed.
Since, X is disconnected by definition, $\exists$ non-empty sets A & B such that $X=A\cup B$, $\bar{A}\cap B=\phi$, $\bar{B}\cap A=\phi$.
Since, $A\neq\phi$,$B\neq\phi$ and $X=A\cup B$, $A\cap B=\phi$,
$\implies A=X\backslash B$.
$\therefore$ A is non-empty proper subset of X.
Also, $B=(\bar{A})^C$, $A=(\bar{B})^C$.
Clearly, A and B are open subset of X.
(Since, complement of closed set is open)
$\therefore A=X\setminus B$ is closed subset of X.
Thus, A is closed as well as open subset of X.
i.e. $\exists$ a non-empty proper set A of X which is both open and closed.
Conversely,
Suppose that metric space (X, d) has non-empty proper subset which is both open and closed.
Claim: X is disconnected.
Suppose A is non-empty proper subset of X which is both open and closed.
Let $B=X\setminus A$
$\therefore B\neq\phi$, $X=A\cup B$ and $A\cap B=\phi$
Since, A is closed and open subset of X, B is open as well as closed.
$\therefore \bar{A}=A$ & $\bar{B}=B$
$\therefore \bar{A}\cap B=A\cap B=\phi$
$A\cap \bar{B}=A\cap B=\phi$
Thus, $X=A\cup B$, $A\neq\phi$, $B\neq\phi$, $\bar{A}\cap B=\phi$, $\bar{B}\cap A=\phi$
$\therefore$ X is disconnected.