Open Set may not be Open Ball

Theorem
Let $M = \left({A, d}\right)$ be a metric space.

Let $U \subseteq M$ be an open set of $M$.

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

Proof
Consider the Euclidean space $\R^2$.

Let:
 * $U \subseteq \R^2: U = \left\{{\left({x_1, x_2}\right): a < x_1 < b, c < x_2 < d}\right\}$

Let $x = \left({x_1, x_2}\right) \in U$.

Then $B_\epsilon \left({x}\right) \subseteq U$ when $\epsilon = \min \left\{{x_1 - a, b - x_1, x_2 - c, d - x_2}\right\}$:


 * NeighborhoodInOpenSet.png

So by definition, $U$ is open in $M$.

However, $U$ is not an open ball.