Let $X$ be a metric space, with $d$ denoting the metric on $X$.
Let $\bar{B} = \bar{B}(a,r)$ denote the closed ball with center at $a \in X$ and radius $r$, i.e., $\bar{B} = \{y \in X : d(y,a) \leq r \}$.
Let $(x_n)_{n\in \mathbb{N}}$ be a sequence of elements in $\bar{B}$ converging to some $x \in X$.
To show that $\bar{B}$ is closed, we can show that $x \in \bar{B}$.
My attempt:
$d(x,a) = d(\lim_{n \to \infty} x_n,a) = \lim_{n \to \infty} d(x_n,a) \leq r$.
The last equality follows from the continuity of the metric function, and the inequality from the fact that $x_n \in \bar{B}, \ \forall n\in \mathbb{N}$.
Thus, $x \in \bar{B}$, and therefore $\bar{B}$ is closed.
My problem: Doesn't this also show that open balls are closed?
Let things be as above, except consider instead the open ball $B = B(a,r)$, i.e., $B = \{y \in X : d(y,a) < r \}$.
Then, arguing as above,
$d(x,a) = d(\lim_{n \to \infty}x_n,a) = \lim_{n \to \infty}d(x_n,a) < r$.
Thus, $x \in B$, and so $B$ is closed.
As I know open balls are open sets, this is clearly wrong.
I think my mistake is that the strict inequality in the last step above does not always hold, as $(x_n)_{n \in \mathbb{N}}$ could converge to a boundary point of $B$.
But $x_n \in B$ for every $n \in \mathbb{N}$, so we must have $d(x_n,a) < r, \forall n \in \mathbb{N}$, no? If I can (perhaps I can't) argue that way with respect to the closed ball, then why can't I argue that way with respect to the open ball?
Yes, the last step doesn't work. $a_n<a$ for all $n$ does not imply $\lim_{n\to\infty} a_n<a$. For example, $a_n=a-1/n$ clearly tends to $a$ despite every term being strictly less than $a$.
Here you have $d(x_n,a)<r$ for every $n$ but that does not imply $\lim_{n\to\infty} d(x_n,a)<r$.