Prove that a closed ball in a metric space is a closed set I attempt to prove this without using the fact that the complement of a closed set is open. I was unable to find this approach on stackexchange.
I would appreciate answers to three questions:
(1) Is the proof correct? (2) Is it preferable to prove using the fact that complements of open sets are closed? If so, why? (3) I realize a lot of questions I ask have already been answered using a different approach on this website. Is it ok for me to ask for a review of alternative approaches such as this?
Proof:
Let $(X,d)$ be a metric space. Let the closed ball be denoted by $\bar{B}_r(x) \subseteq X$. We need to show that all limit points for any sequence $\{b_n\} \subseteq \bar{B}_r(x)$ is contained in $\bar{B}_r(x)$. For sake of contradiction, assume there exists a limit point, $L$ for $\{b_n\}$, such that $L \not\subseteq \bar{B}_r(x)$. Let $b \in \bar{B}_r(x)$ be the point of the closed ball that is closest to $L$, and let $\epsilon=\frac{d(b,L)}{2}$. By construction, a ball $B_\epsilon(L)$ does not overlap with $\bar{B}_r(x)$. However, by convergence, $B_\epsilon(L)$ must contain all but finitely many points of the convergent sequence $\{b_n\} \in \bar{B}_r(x)$. We thus, have that $\bar{B}_\epsilon(L)$ contains infinitely many points of $\bar{B}_r(x)$, but $B_\epsilon(L) \not\subseteq \bar{B}_r(x)$. This is a contradiction. This shows that if $L$ is a limit point of a closed ball, it must be contained within the closed ball $\implies$ a closed ball is a closed set. 
 A: Looks like a good proof to me. You assume L is a limit point of the ball but is not inside the ball. In order for it not to be in the ball, there must be some $\varepsilon$ ball around it which is also not in the closed ball, but since L is a limit point there must be some sequence in the closed ball which converges to L. This means that points in the closed ball can be arbitrarily close to L, but this is a contradiction.
A: [1].Let $(X,d)$ be a metric space. A basic useful fact is that if $(x_n)_n$ is a sequence in $X$ converging to $x\in X,$ i.e. if $d(x_n,x)\to 0,$ then $d(y,x_n)\to d(y,x)$ for any $y\in X.$ Proof:
(i). $d(y,x_n)\le d(y,x)+d(x,x_n).$ So $\lim\sup_{n\to\infty}d(y,x_n)\le \lim\sup_{n\to\infty}d(y,x)+d(x,x_n)=d(y,x).$
(ii). $d(y,x_n)\ge d(y,x)-d(x,x_n)$. So $\lim\inf_{n\to\infty}d(y,x_n)\ge \lim\inf_{n\to\infty}d(y,x)-d(x,x_n)=d(y,x ).$
[2]. With $\bar B_r(x)=\{y\in x:d(x,y)\le r\},$ suppose $(y_n)_n$ is a sequence in $\bar B_r(x)$ converging to $y\in X.$ Then $d(y_n,x)$ converges to $d(y,x)$ by [1]. But the set $\{d(y_n,x):n\in \Bbb N\}$ is a subset of the real interval $[0,r]$ so $d(y_n,x)$ converges to a member of $[0,r]$. So $d(y,x)\le r$, that is, $ y\in\bar B_r(x).$
