Understanding a proof about a set being closed 
Let $r > 0$ be a positive number, and define $F = \{u \in
 \mathbb{R}^{n} \mid \|u\| \leq r\}$. Prove $F$ is closed in
  $\mathbb{R}^{n}$.

Proof:
We want to show that if a sequence $\{u_{k}\}$ lies in $F$ and $\lim_{k\to\infty} u_{k} = u$, then $u \in F$. Let $\{u_{k}\}$ be in arbitrary sequence in $F$. By the set definition of $F$, it follows that $\|u_{k}\| \leq r$ for each index $k$. Thus,
$$\lim_{k\to\infty} \|u_{k}\| \leq \lim_{k\to\infty} r = r.$$ 
From here, it suffices to show $\lim_{k\to\infty} \|u_{k}\| = \|u\|$. 
The proof goes on and shows that $\lim_{k\to\infty} \|u_{k}\| = \|u\|$

My question:
I don't understand why it suffices to show that the sequence of norms converges to the norm of the limit point? After they prove this fact, they say that $\lim_{k\to\infty} \|u_{k}\| = \|u\| = r$, from which it follows that $u \in F$, hence $F$ is closed. But, I don't get why proving this shows that $u$ is contained in $F$.
 A: I think you are a bit confused so I'll try to make up the argument with more details. First, showing that a set is closed is equivalent to showing that every converging sequence has a limit point IN the set. Here the set we are concerned with is $F=\{u\in \mathbb{R}^n : ||u||\leq r\}$. Thus, if we pick a sequence $(u_k)_k\subset F$ such that $u_k\to u$, it suffices to show that $u\in F$ to prove that $F$ is closed. What does it mean to be an element of $F$ ? It means simply that $||u||\leq r$ ! So we have to show that $||u||\leq r$ :) That's why it is sufficient to show that the norm of the $u_k$'s goes to the norm of $u$ because $||u_k||\leq r$ for all $k$ and by taking the limit inside the norm because the norm is continuous, we get $||u||\leq r$ which is the condition an element of $\mathbb{R}^n$ has to satisfy to be in $F$.
If you want more details, let me know !
A: The proof provided by Malik is correct.  I'm giving a separate answer because there's another approach that often works in general and is often easier.  A set is closed if its complement is open.  So assume you have a point $x \notin F$, so that $||x|| \gt r$.  Can you prove that there's necessarily an open ball around $r$ that does not intersect $F$?
Hint:  Let $||x|| = r+a, a \gt 0$.  Let $\epsilon = a/2$ and consider $B(x, \epsilon)$ using the triangle inequality.
A: You can also notice that proving
$$
\lim_{k\to \infty} ||u_k||=||u||
$$
proves that $f(u) =||u||$ is continuous. Now,
$$ 
F=\{ u\in \mathbb R^n : ||u||\le r\} =f^{\leftarrow}([0,r]). 
$$
Recalling that if $f$ is continuous and $C$ is closed then $f^{\leftarrow}(C)$ is closed, the result follows from the fact that $[0,r]$ is closed in $\mathbb R$. 
