ProofWiki:Sandbox

Proof
Suppose $d(x,y)$ only can assume at most two different values for $x \ne y$.

That is, there exist $d_1, d_2 \in ]0,\infty$ with $d_1 \le d_2$, such that $d(x,y) \in \{ d_1 , d_2 \}$ for all $x , y \in M$ with $x \ne y$.

Find two distinct points $x,y \in M$, such that $d(x,y) = d_2$.

Set $U = B( x; \frac{ d_1 }{ 2 } ) \cup B( y; \frac{ d_1 }{ 2 } ) = \{ x, y \} $, so $U$ is open.

Let $a \in M$ and $\epsilon > 0 $.

If $a \notin M$, we have $B( a; \epsilon ) \ne U$, as $a \in B( a; \epsilon )$.

If $a = x$, we have $y \notin B( a; \epsilon )$ if $\epsilon < d_2$, and $B( a; \epsilon ) = M$ if $\epsilon \ge d_2$.

In both cases, we have $B( a; \epsilon ) \ne U$.

A symmetry argument shows that if $a = y$, we have $B( a; \epsilon ) \ne U$.