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 $x, y, z \in A$ be $3$ distinct points such that $d \left({x, y}\right) \ge d \left({x, z}\right)$ and $d \left({x, y}\right) \ge d \left({y, z}\right)$.

Put $\epsilon_0 = \min \left({\frac 1 2 d \left({x, y}\right), d \left({x, z}\right), d \left({y, z}\right)}\right)$.

Put $U_0 = B_x \left({\epsilon_0}\right) \cup B_y \left({\epsilon_0}\right)$.

From Union of Open Sets of Metric Space, $U_0$ is open.

From Distinct Points in Metric Space have Disjoint Open Balls, it follows that the open balls $B_x \left({\epsilon_0}\right)$ and $B_y \left({\epsilon_0}\right)$ are pairwise disjoint.

As $\epsilon_0 \le \min \left({d \left({x, z}\right), d \left({y, z}\right)}\right)$, it follows that $z \notin U_0$.

Suppose that $\exists a_0 \in A, \epsilon \in \R_{>0}$ such that $U_0 = B_{a_0} \left({\epsilon}\right)$, otherwise $U_0$ would fulfill the requirement.

Then either $a_0 \in B_x \left({\epsilon_0}\right)$ or $a_0 \in B_y \left({\epsilon_0}\right)$.

WLOG assume that $a_0 \in B_x \left({\epsilon_0}\right)$.

Suppose $a_0 = x$.

As $y \in U_0$, it follows that $\epsilon \ge d \left({x, y}\right) \ge d \left({x, z}\right)$.

Then $z \in U_0$ which is a contradiction.

Hence, $a_0 \ne x$, and so $d \left({x, a_0}\right) > 0$.

Put $\epsilon_1 = \frac 1 2 d \left({x, a_0}\right) < \epsilon_0$.

Put $U_1 = B_x \left({\epsilon_1}\right) \cup B_y \left({\epsilon_0}\right)$.

Then $a_0 \notin U_1$, and $U_1$ is open by Union of Open Sets of Metric Space.

Suppose that $\exists a_1 \in A, \epsilon \in \R_{>0}$ such that $U_1 = B_{a_1} \left({\epsilon_0}\right)$, otherwise $U_1$ would fulfill the requirement.

Then either $a_1 \in B_x \left({\epsilon_1}\right)$ or $a_1 \in B_y \left({\epsilon_0}\right)$.

Suppose that $a_1 \in B_x \left({\epsilon_1}\right)$.

Then:

As $y \in U_1$, it follows that $\epsilon > d \left({a_1, y}\right) > d \left({a_1, a_0}\right)$.

So $a_0 \in U_1$, which is a contradiction.

Hence, $a_1 \notin B_x \left({\epsilon_1}\right)$, so $a_1 \in B_y \left({\epsilon_0}\right)$.

Then $a_1 \ne y$, otherwise $z \in B_y \left({\epsilon}\right)$, as has been noted before with a similar argument.

Put $\epsilon_2 = \frac 1 2 d \left({y, a_1}\right) < \epsilon_0$.

Put $U_2 = B_x \left({\epsilon_1}\right) \cup B_y \left({\epsilon_2}\right)$.

Then $a_0 \notin U_2, a_1 \notin U_2$, and $U_2$ is open.

Suppose that $\exists a_2 \in A, \epsilon \in \R_{>0}$ such that $U_2 = B_{a_2} \left({\epsilon}\right)$.

Suppose that $a_2 \in B_x \left({\epsilon_1}\right)$ with $y \in B_{a_2} \left({\epsilon}\right)$.

Then, it follows as above that $a_0 \in B_{a_2} \left({\epsilon}\right)$, which is a contradiction.

Suppose instead that $a_2 \in B_y \left({\epsilon_2}\right)$ with $x \in B_{a_2} \left({\epsilon}\right)$.

A similar argument shows that $a_1 \in B_{a_2} \left({\epsilon}\right)$, which is a contradiction.

Hence, it is impossible that $U_2 = B_{a_2} \left({\epsilon}\right)$.

It follows that either $U_2$, $U_1$ or $U_0$ is an open set in $M$ which is not an open ball.