Open Set may not be Open Ball

Theorem
Let $M = \left({A, d}\right)$ be a metric space with at least $3$ distinct points.

Let $U \subseteq A$ be a non-empty open set of $M$.

Then it is not necessarily the case that $U$ is an open ball of some $x \in A$.

That is, there exists $U \subseteq A$ which is open in $M$ which is not an open ball.

Proof
Let $x, y, z \in U$ such that $d \left({x, y}\right) \ge d \left({x, z}\right)$ and $d \left({x, y}\right) \ge d \left({y, z}\right)$.

From Distinct Points in Metric Space have Disjoint Open Balls we can find open balls $B \left({x; \epsilon_x}\right)$, $B \left({y; \epsilon_y}\right)$ and $B \left({z; \epsilon_z}\right)$ which are pairwise disjoint.

From Union of Open Subsets, $B \left({x; \epsilon_x}\right) \cup B \left({y; \epsilon_y}\right)$ is an open set of $M$ which is non-empty as it contains $x$ and $y$.

Now suppose $B \left({x; \epsilon_x}\right) \cup B \left({y; \epsilon_y}\right)$ were an open ball.