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:) 

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.
