Distinct Points in Metric Space have Disjoint Neighborhoods

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

Let $x, y \in M: x \ne y$.

Then there exist neighborhoods $N_x$ and $N_y$ of $x$ and $y$ respectively such that $N_x \cap N_y = \varnothing$, that is, that are disjoint.

Proof
Let $x, y \in A: x \ne y$.

From Distinct Points in Metric Space have Disjoint Open Balls, there exist disjoint open $\epsilon$-balls $B_\epsilon \left({x}\right)$ and $B_\epsilon \left({y}\right)$ containing $x$ and $y$ respectively.

From Open Ball is Neighborhood of all Points Inside it follows that $B_\epsilon \left({x}\right)$ and $B_\epsilon \left({y}\right)$ are neighborhoods of $x$ and $y$ respectively.

Hence the result.