The idea of this proof is quite clear but I'm having some trouble making it rigorous. Suppose we have a metric space $(X, d)$ and a closed ball $U := \{x \in X : d(x, a) \leq t\}$ for some fixed $a$ and $t$. I want to prove that a point on the boundary of this ball is not an interior point. Here is my "proof":
Let $x$ satisfy $d(x, a) = t$ (i.e. let $x$ be a boundary point). Suppose also that $x$ is interior. Then $\exists \, r > 0$ such that the open ball $D_r(x)$ is contained within $U$. This an immediate contradiction, because some points in this open ball are outside $U$.
My problem is with the very last statement, which relies entirely upon geometrical intuition and is not very rigorous. I suppose I could try a bit harder with this idea: along the line connecting $a$ and $x$, we can go a bit further along the line still inside the $r$ -ball and find a point outside of $U$. But this still doesn't sound very rigorous, with things like lines only really applying to Euclidean space.
How can I make this rigorous?
EDIT: Thanks for the answers and comments, I now realize that this cannot be proven at all.
Answer
In a general metric space the boundary of the set $U = \{x : d(x,a) \le t\}$ is not the set $\{x : d(x,a) = t\}$.
The (usual) definition of boundary point of a set implies that the boundary and interior of a set are disjoint.
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