Extremally Disconnected Metric Space is Discrete

Theorem
Let $M = \left({A, d}\right)$ be a metric space which is extremally disconnected.

Then $M$ is the discrete topology.

Proof
Let $M = \left({A, d}\right)$ be extremally disconnected.

Let $p \in A$.

As $M$ is a metric space, $\left\{{p}\right\}$ can be expressed as:
 * $\left\{{p}\right\} = \displaystyle \bigcap_{n \mathop \in \N_{>0}} \left({B_{1 / n} \left({p}\right)}\right)^-$

where:
 * $B_{1 / n} \left({p}\right)$ denotes the open $1 / n$-ball of $p$
 * $\left({B_{1 / n} \left({p}\right)}\right)^-$ denotes the closure of $B_{1 / n} \left({p}\right)$

That is, as the intersection of the closures of the open $1 / n$-ball of $p$ for all non-zero natural numbers.

Now let:
 * $\displaystyle U = \bigcup_{n \mathop \in \N_{>0}} B_{1 / 2 n} \left({p}\right) \setminus \left({B_{1 / \left({2 n + 1}\right)} \left({p}\right)}\right)^-$

Then either $U$ or the complementary set of annuli is an open set which has $p$ as a non-interior limit point provided $\left\{{p}\right\}$ is not open.

So if $M$ is not the discrete topology, it cannot be extremally disconnected.