Open Set may not be Open Ball

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

Then there exists $U \subseteq A$, where $U$ is open in $M$, but is not an open ball.

Proof
We will construct three candidates for $U$, and show that at least one of these candidates is an open set that is not an open ball.

Construction of first candidate for $U$
Let $x, y, z \in A$ be three distinct elements 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 $\epsilon_0 = \min \left({\dfrac 1 2 d \left({x, y}\right), d \left({x, z}\right), d \left({y, z}\right)}\right)$.

Let $B_x \left({\epsilon_0}\right)$ denote the open ball of $x$ with radius $\epsilon_0$.

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 $z \notin U_0$.

Construction of second candidate for $U$
If $U_0$ is not an open ball, then $U_0$ is open in $M$ and is not an open ball.

Otherwise there exists $a_0 \in A$ and $\epsilon \in \R_{>0}$ such that $U_0 = B_{a_0} \left({\epsilon}\right)$.

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 $\epsilon > d \left({x, y}\right) \ge d \left({x, z}\right)$.

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

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

Put $\epsilon_1 = \dfrac 1 2 d \left({x, a_0}\right) < \dfrac 1 2 \epsilon_0 \le \dfrac 1 4 d \left({x, y}\right)$.

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.

Construction of third candidate for $U$
If $U_1$ is not an open ball, then $U_1$ is open in $M$ and is not an open ball.

Otherwise there exists $a_1 \in A$ and $\epsilon \in \R_{>0}$ such that $U_1 = B_{a_1} \left({\epsilon_0}\right)$.

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

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

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.

It follows that $a_1 \notin B_x \left({\epsilon_1}\right)$, and so $a_1 \in B_y \left({\epsilon_0}\right)$.

Then $a_1 \ne y$, otherwise $z \in B_{ a_1 } \left({\epsilon}\right)$, as has been noted with a similar argument during the construction of $U_1$.

Put $\epsilon_2 = \dfrac 1 2 d \left({y, a_1}\right) < \dfrac 1 2 \epsilon_0 \le \dfrac 1 4 d \left({x, y}\right)$.

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, to prove a contradiction, that there exists $a_2 \in A$ and $\epsilon \in \R_{>0}$ 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.