Prove that every closed ball in $\Bbb R^n$ is sequentially compact. Question:
Prove that every closed ball in $\Bbb R^n$ is sequentially compact. 

A subset $E$ of $\Bbb R^n$ is said to be squentially compact $\iff$ every sequence $x_k\in E$ has convergent subsequence whose limit belongs to $E$ 

Solution: 
Let $B_R(a)$ be closed ball. Let $x_k$ be a sequence in $B_R(a)$ 
Then, $$\vert\vert x_k-a\vert\vert \le M$$ for $M>0$ 
By the triangle inequality, 
$$\vert\vert x_k-a\vert\vert \le \vert \vert x_k\vert \vert +\vert\vert a\vert \le M \ \  \Rightarrow \vert\vert x_k\vert\vert \le \vert\vert a\vert\vert +M $$ 
So the sequence $x_k$ is bounded. 
By Bolzano W. Theorem, $x_k$ has convergent subsequences. 

Now I need to show that these convergent subsequences have a limit point in $B_R(a)$. 
But how? Please explain this part. Thank you:) 
 A: What you wrote is not entirely correct. What you should've used is the reverse triangle inequality $$|\lVert x_k\rVert-\lVert a\rVert |\leq \lVert x_k-a\rVert\leq M$$
This gives that $$-M\leq\lVert x_k\rVert-\lVert a\rVert\leq M$$ $$\lVert a\rVert-M\leq\lVert x_k\rVert\leq M+\lVert a\rVert$$
so the sequence is bounded. It seems strange to use Bolzano Weiertrass, since what you're being asked to prove is precisely that theorem. I would argue as follows. You probably know the result of Bolzano Weiertrass for $\Bbb R^1$. But one can extend it easily to $\Bbb R^n$. Consider a bounded sequence in $\Bbb R^n$ $${\bf x}_k=(x_{k,1},x_{k,2},\ldots,x_{k,n})$$
By Bolzano Weiertrass, $x_{k,1}$ has convergent subsequence, $y_{k,1}=x_{n_k,1}$. Now look at $${\bf x}_{n_k}={\bf y}_k=(y_{k,1},y_{k,2},\ldots,y_{k,n})$$
This is now a subsequence of the whole ${\bf x}_k$. We know the subsequence $y_{k,2}=x_{n_k,2}$ is bounded, so it has a convergent subsequence by Bolzano Weiertrass in $\Bbb R^1$, which we will call $z_{k,2}=y_{k_j,2}$. Now we have a subsequence of ${\bf y}_k$ (which was already a subsequence of ${\bf x}$, whose two first coordinates have matching subindices and both converge. And as you should see by now, we ultimately obtain a subsequence of "deepness $n$" (meaning it will be the subsequence of a subsequence of ... a subsequence of ${\bf x}$ - $n$ times) which we can call $${\bf x'}_j=(x_{j,1}^\prime,\ldots, x_{j,n}^\prime)$$ which converges, since each coordinate converges. Thus, Bolzano Weiertrass is proven for $\Bbb R^n$. Since closed balls are closed sets, your claim follows, since closed sets contain their limit points.

NOTE Bolzano Weiertrass is precisely the statement that every closed ball in $\Bbb R^n$ is compact, equivalently sequentially compact, or that $\Bbb R^n$ is locally compact, that is, every point $a\in\Bbb R^n$ has a neighborhood whose closure is compact.

A: First, note that a closed ball is a closed set. 
To prove this, let $E\subset \mathbb{R}$ is a closed ball of radius $R$ around $y$. Let $x\in E^{C}$. Now, let $r=d(x,y)-R>0$ and $z\in B_{r}(x)$. Then $$d(y,x)\le d(y,z)+d(z,x)<d(y,z)+r\\
\Rightarrow d(y,z)>R$$ Hence $z\notin E$ and hence the open ball $B_r(x)$ is totally contained in $E^{C}$. hence $E^{C}$ is open and hence $E$ is closed.
Now, since you've got a convergent (sub)sequence in a closed set, it is going to converge to some point in the set itself, by definition of closed set. So you're done.
