Show that, in an infinite dimensional normed space $(V,\|\cdot\|)$, the closed ball of radius $2$ $$ B_2:=\{x\in V:\ \|x\|\leq2\} $$ is not compact.
I suspect I am not understanding what is going on and have no idea where the proof whould start from.
On
Hint: First of all, try to prove F.Ries's Lemma: Let $Y$ and $Z$ be subspaces of a normed space $X$, and suppose that $Y$ is closed and is a proper subsets of $Z$. Then for every number $\theta$ in the interval $(0;1)$ there is a $z \in Z$ such that $ \left \|z \right \|=1$, and $\left \|z-y \right \| \geq \theta$, $\forall y \in Y$.
Using the above lemma it suffices to prove that $B=\left \{x \mid \left \|x \right \| \leq1 \right \}$ is not compact.
Assume a contradiction that $B$ is compact. We choose any $x_i$ of norm $1$. This $x_i$ generates a one dimensional subspace $V_i$ of $V$, which is closed and is a proper subspace of $V$ since $dimV = \infty$. By Riesz's lemma there is an $x_2 \in V$ of norm $1$ such that $$ \left \|x_2 -x_1 \right \| \geq\theta=\frac{1}{2} $$
The elements $x_1, x_2$ generate a two dimensional proper closed subspace $V_2$ of $V$. By Riesz's lemma there is an $x_3$ of norm $1$ such that for all $x\in V_2$ we have: $$ \left \|x_3 -x \right \| \geq\frac{1}{2} $$
In particular,$$ \left \|x_3 -x_1 \right \| \geq\frac{1}{2}, \left \|x_3 -x_2 \right \| \geq\frac{1}{2} $$
Proceeding by induction, we obtain a sequence $(x_n)$ of elements $x_n \in B$ such that $$ \left \|x_m -x_n \right \| \geq\frac{1}{2} $$
Obviously, $(x_n)$ cannot have a convergent subsequence. This
contradicts the compactness of $B$. Consequently, $B$ is non-compact, as desired.
Hint: As $V$ is a normed space, it is also a metric space. In a metric space, compactness is equivalent to sequential compactness. Can you think of a sequence in $B_2$ which has no convergent subsequence? Note, as the result is false if $V$ is finite dimensional, the sequence must take advantage of the fact that $V$ is infinite dimensional.