TITLE: Cardinality of an infinite separable connected metric space is $2^{\aleph_0}$.
QUESTION [5 upvotes]: How to prove:
Cardinality of an infinite separable connected metric space is $2^{\aleph_0}$.
Thanks in advance!!

REPLY [4 votes]: To see that $|X| \ge 2^{\aleph_0}$, we don't need to use complete regularity.  Pick $x \in X$; I claim that for every sufficiently small $r$, say $r < R$, there exists $x_r \in X$ with $d(x, x_r) = r$.  (Suppose not; then you can show that $B(x,r)$ is open and closed.)  Since the points $\{x_r\}$ are necessarily distinct, you have a set of the cardinality of $(0,R)$.