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Proof of Open Set may not be Open Ball
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 $, so $ d(x,y) = d_1 $ or $ d(x,y) = d_2 $ for all $ x , y \in M $ with $ x \ne y $.

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

Let $ U = B_{ d_1 / 2 }( x ) \cup B_{ d_1 / 2 }( y ) = \{ x, y \} $, so $ U $ is open.

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

If $ a \notin M $, we have $ B_{ \epsilon }( a ) \ne U $, as $ a \in B_{ \epsilon }( a ) $.

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

In both cases, we have $ B_{ \epsilon }( a ) \ne U $.

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

Suppose instead that $ d(x,y) $ can assume more than two different values for $ x \ne y $.

Then we can find three distinct points $ x, y, z \in M $, so $ d(x,y) > d(x,z) $, and $ d(x,y) > d(y,z) $.

Let $ r = \min ( \frac{ d(x,y) - d(x,z) }{ 2 }, d(x,z) ) $, and $ s = \min ( \frac{ d(x,y) - d(y,z) }{ 2 } , d(y,z) ) $.

Let $ U = B_r ( x ) \cup B_s ( y ) $.

Then $ U $ is open, and $ z \notin U $.

If $ a \notin M $, we have $ B_{ \epsilon } ( a ) \ne U $, as $ a \in B_{ \epsilon ]( a ) $.

If $ a \in M $, then $ a \in B_r ( x ) $ or $ a \in B_s ( y ) $.

Suppose $ a \in B_x $, then

{{eqn | l= > | d(x,y) - \frac{ d(x,y) - d(x,z) }{ 2 } | r= \mathrm{ as } d(x,a) < r

{{end-eqn}}