Is it true that for any open ball $B(x;r)$, its boundary must be the sphere, $S(x;r)=\{y \in X : d(x,y)=r\}$?
So I am trying to go about this with initial intuition. Since an open ball is defined by $d(x,y)<r$, it must be that the closure is indeed going to be the points on the sphere, but this doesn't seem too convincing to me.
Are my thoughts correct or am I missing something?
No. Consider the discrete topology, then every set is open and closed and so the boundary is the empty set.
Edit: You can define a metric $ d(x,y) = 1 $ if $x\neq y$ and $ d(x,y) = 0 $ if $ x = y $. So if $ (X,d) $ is a metric space, then you can show that every set is open and closed. Thus $ U \subset X$ is open so the boundary is in $ X - U $ and likewise $ X - U $ is open so the boundary is in $ U $ but $ (X - U) \cap U = \emptyset $. So the boundary must be the empty set.
Infimum is right with their counterexample, but let me offer non-trivial insight.
The boundary of an open ball $O \subseteq S$ is precisely $\overline{O} - O$. Why? From Wikipedia, the boundary of a set $S$ is..
...the set of points in the closure of S, not belonging to the interior of S.
Since $O$ is open, its interior points are exactly $O$, and so $\overline O - O$ is the boundary of $O$.
You have asked when it is the case that: \begin{align} \partial B(x;r) &= \{y \in M : d(x,y) = r\} \\ &= \{y \in M : d(x,y) \leq r\} - \{y \in M : d(x,y) < r\} \\ &= B_{closed}(x;r) - B(x;r) \\ \overline{B(x;r)} - B(x;r) &= B_{closed}(x;r) - B(x;r) \end{align}
It is obvious that both $\overline{B(x;r)}$ and $B_{closed}(x;r)$ contain the entirity of $B(x;r)$, so we may reduce this to: $$\boxed{\overline{B(x;r)} = B_{closed}(x;r)} \quad \Leftrightarrow \quad \partial B(x;r) = \{y \in M : d(x,y) = r\}$$
Knowing this, your question becomes equivalent to the following:
When is it the case that the closure of an open ball is equal to the closed ball?
...which is answered here with the following equivalent condition:
For any two distinct points $x,y$ in the space and any positive $\epsilon$, there is a point $z$ within $\epsilon$ of $y$, and closer to $x$ than $y$ is. That is, for every $x\neq y$ and $\epsilon\gt 0$, there is $z$ with $d(z,y)<\epsilon$ and $d(x,z)<d(x,y)$.
