If we have ball $X$ is compact. How to get the set $\{x\in X:\|x\|\leq M\}$ compact? $M$ is a constant number Let $X$ be a normed space. Let ball $X$ be a closed unit ball of $X$. We have ball $X$ is compact. How to get the set $\{x\in X:\|x\|\leq M\}$ compact? $M$ is a constant number.
 A: Hint Let $B$ be the unit ball.
Show that 
$$F : B \times [0,M] \to \{x\in X:||x||\leq M \} \\
F(x, t)=tx$$
is continuous.
A: For 'cover compact' you can argue as follows:
Let $B$ denote a closed subset of $X$. Then $X\backslash B$ is an open set in $X$.
Let $\mathcal{U}$ be a collection of open sets covering $B$ and let $\mathcal{U}^\prime=\mathcal{U}\cup(X\backslash B)$. 
Then $\mathcal{U}^\prime$ covers $X$ so some finite subcollection $\mathcal{V}\subseteq\mathcal{U}^\prime\backslash (X\backslash B)\subseteq\mathcal{U}$ covers $B$.
A: Since the norm topology is given by a metric, we have that compactness is equivalent to sequential compactness. Thus to show that $\{x\in X \mid \|x\|\leq M\}$ is compact, it suffices to show that every sequence in it has a convergent subsequence. Suppose $(x_n)_n$ is a sequence such that $\|x_n\|\leq M$. If $y_n:=x_n/M$, then $(y_n)_n$ is a sequence in the unit ball of $X$. It follows from the compactness of the unit ball that there is $y$ such that $\|y\|\leq 1$ and $y_{n_k}\to y$ for some subsequence $(y_{n_k})_k$. Then $x:=My$ satisifies $\|x\|\leq M$ and
$$
\|x_{n_k}-x\| = \|My_{n_k} - My\| = M\|y_{n_k}-y\|\to 0.
$$
