“Every open ball is closed" and "every closed ball is open", does one imply the other? In a metric space $(X,d)$, by a closed ball I mean a set of the form $\{y: d(y,x)\le r\}$, where $r > 0$.
A common example where every open ball is a closed set and every closed ball is an open set is the ultrametric spaces: the metric spaces where $d(x,y)\le \max\{d(x,z),d(y,z)\}$. I would like to have an example where every open ball is a closed set but not every closed ball is an open set, and an example vice versa.

Definitions.  An open ball is a set of the form $B_r(a) = \{x \mid d(x,a) < r\}$ where $r > 0$.  A closed ball  is a set of the form $\overline{B}_r(a) = \{x \mid d(x,a) \le r\}$ where $r > 0$.
 A: Here's an example going the other direction: all open balls are closed, but there a closed ball which is not open.  But let me note that I'm going to use the notation $CB_r(p)$ for the closed ball, rather than $\overline{B}_r(p)$, because the latter notation makes me think of the closure of an open ball.  Then one has $\overline{B}_r(p)\subseteq CB_r(p)$, but the inclusion can be strict.  E.g., in a discrete metric space $\overline{B}_1(p) = p$, while $CB_1(p)$ is the whole space.
Consider the Hilbert cube $H:=[0,1]^\mathbb{N}$ with metric $d( (x_1,x_2,...), (y_1,y_2,..)) = \sum_i \frac{|x_i - y_i|}{2^i}$.
For each $i\in \mathbb{N}$, let $e_i$ denote the element of $H$ which is all $0$s except for a $1$ in the $i$th slot.  So, $e_1 = (1,0,0,..)$, $e_2  = (0,1,0,...)$, etc.  I'll also write $0$ for the all-zeroes element $(0,0,...)$
Let $Y\subseteq H$ be given by $Y = \{e_i, 0\}$.  Then $Y$ will be the desired example.  That is, I claim that 1)  every open ball in $Y$ is closed but  2) that there is a closed ball in $Y$ which is not open.
As a preliminary observation, note that for $i\neq j$, that $d(e_i,e_j) = 1/2^i + 1/2^j > \max\{1/2^i, 1/2^j\}$, and that $d(e_i,0) = 1/2^i$.  It follows that each $e_i$ is isolated, with, e.g., $B_{1/2^i}(e_i) = \{e_i\}$ being an open ball containing just the one point $e_i$.  Moreover, for each $e_i$, $0$ is the unique closest point to it.
Here's the proof of 1)  To begin with, note first that every open ball centered at $0$ is closed because the complement can only possible contain some of the $\{e_i\}$ which are all isolated.
So, consider $B = B_{r}(e_i)$, an open ball centered at $e_i$ of radius $r > 0$.  Since all the $e_i$ are isolated, the only case we need to consider is if $0\notin B$.  But $0$ is the closest point to $e_i$, so if $0$ is not in $B$, then neither are any $e_j$ with $i\neq j$.  Thus, any ball around $0$ which doesn't contain $e_i$ witnesses the fact that the complement of $B$ is open.  E.g., one can take $B_{1/2^i}(0)$.  This concludes the proof of 1).
We now prove 2), that there is a closed ball which is not open.  To that end, consider $C:=CB_{1/2}(e_1)$.  This set clearly contains $e_1$ and $0$; since $0$ is the unique closest point to $e_1$, it follows that $C = \{e_1,0\}$.  So, to show $C$ is not open, it's enough to show that every ball around $0$ contains an $e_i$ other than $e_1$.
But $d(e_i,0) = 1/2^i\rightarrow 0$ as $i\rightarrow \infty$, so any open ball around $0$ contains all but finitely many of the $e_i$.
A: Example 1.  In the line, consider the set
$X = \{0, 1, 1/2, 1/3, 1/4,\dots\}$ with the usual metric.  I claim:
every closed ball in $X$ is open in $X$; but the open ball ${B}_1(1)$ is not closed in $X$.
