# Proof :

Let (X, d) be a connected metric space and suppose that there exists a continuous function $f:X\to {\lbrace0, 1\rbrace}$.
Claim: f is constant.
Since, ${\lbrace0, 1\rbrace}$ is finite w.r.t. usual metric, set ${\lbrace 0\rbrace}$ and $\lbrace 1 \rbrace$ are open and closed subsets of $\lbrace 0, 1 \rbrace$,
$\therefore f^{-1}\lbrace 0 \rbrace$ and $f^{-1}\lbrace 1 \rbrace$ both are open and closed subset of X. ($\because$ f is continuous)
$f^{-1}{\lbrace 0 \rbrace} \cup f^{-1}{\lbrace 1\rbrace}=X$ and $f^{-1}{\lbrace 0 \rbrace} \cap f^{-1}{\lbrace 1\rbrace}=\phi$.
Thus, X is union of disjoint sets.
But, by hypothesis,
X is connected and hence, either $f^{-1}\lbrace 0 \rbrace=\phi$ or $f^{-1}\lbrace 1 \rbrace=\phi$.
WLOG,
Suppose, $f^{-1}\lbrace 0 \rbrace=\phi$.
$\therefore f^{-1}\lbrace 1 \rbrace=X$
$\therefore f(X)=1$
i.e. f is constant.
Conversely,
Suppose that every continuous function $f:X\to {\lbrace0, 1\rbrace}$ is constant.
Claim: Metric space (X, d) is connected.
Since, $f:X\to {\lbrace0, 1\rbrace}$ is constant function,
either $f(X)=\lbrace 0 \rbrace$ or $f(X)=\lbrace 1 \rbrace$.
WLOG, suppose that $f(X)=\lbrace 1 \rbrace$.
Since, $\lbrace 1 \rbrace$ is closed as well as open subset of ${\lbrace0, 1\rbrace}$ w.r.t. usual metric and f is continuous function, $f^{-1}\lbrace 1 \rbrace$ is closed as well as open.
But $f^{-1}\lbrace 1 \rbrace=X$.
$\therefore$ X is only nonempty set which is both open as well as closed in X.
$\therefore$ X is connected.

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