Closed balls are closed Let $E \subset X, p \in X$ and $(X,d)$ a metric space such that $E := \{q \in X\mid d(p,q) \leq r\}$. By definition, this is a closed ball with center $p$ and radius $r$. I now want to show that this set is closed (i.e. it contains all its limit points).
Proof
Let $k$ be a limit point of $E$. Let $\epsilon >0$ and define $N_{\epsilon}(k)$ as the open ball (neighborhood) with center $k$ and radius $\epsilon$. By definition of limit point, there exists $z \in E \cap N_{\epsilon}(k)$ such that $z \neq k$. Hence we have $d(p,z) \leq r$ and $d(k,z) < \epsilon$, so by triangle inequality, we have $d(p,k) < r + \epsilon$. Because $\epsilon >0$ is chosen arbitrarily, it follows that $d(p,k) \leq r$ (if $d(p,k) > r$, then $d(p,k) - r > 0$, implying that $d(p,k) < d(p,k)$, which is absurd). We deduce that $k \in E$ and hence $E$ is closed. $\quad \square$
Is my proof correct? Is there an easier proof?
 A: Your proof is correct.You proved that the closure of $E$ is equal to $E$ thus $E$ is closed.
To answer also your second question,there is  an easier proof for this:
Let $x_n \in E$ such that $x_n \rightarrow x$.
Then $d(x,p) \leq d(x,x_n)+d(x_n,p) \leq d(x,x_n)+r$
Taking limits to both sides of the inequality we have that 
$$d(x,p) \leq r \Rightarrow x \in E$$
Thus $E$ is closed.
A: An alternate proof is to show that $X - E$ is open. To show that $x \in X - E$ is an interior point, consider the open ball centered at $x$ with radius $d(x, p) - 1$, and then argue that this is disjoint from $E$ using the triangle inequality.
A: Your proof looks good to me.  The key point, that $a<b+\epsilon$ for all $\epsilon>0$ implies $a\le b,$ is used explicitly, which is good expository style.
An alternative proof: the "closed" ball $E$ is the inverse image of the closed set $[0,r]$ under the continuous map $q\mapsto d(p,q).$   Hence it is closed.  Or, it is the complement of the inverse image of the open set $(r,\infty)$, and thus closed.
