Sow that if $\mathcal{H}$ is not finite-dimensional, then there exists a sequence in $B$ that does not have a convergent subsequence.

Consider a Hilbert space $\mathcal{H}$ with inner product $\langle \cdot, \cdot \rangle$ and norm $|| \cdot ||$. Consider the set

$$B = \{x \in \mathcal{H} | ||x|| \leq 1 \} $$

Then I have to show that if $H$ is not finite-dimensional, then there exists a sequence in $B$ that does not have a convergent subsequence.

My attempt:

Consider the orthonormal system $\{e_n : n \in \mathbb{N}\}$ in $\mathcal{H}$ and $\{e_m : m \in \mathbb{N}\}$ in $\mathcal{H}$ where $e_n \perp e_m$ and $e_n,e_m \in B$ as $||e_n||=||e_m||=1$. We know that a sequence is convergent if and only if it is cauchy. Therefore, it is sufficient to show that the sequence $e_n$ is not cauchy.

As $e_n \perp e_m$ it follows from Pythagoras that $$||e_n + (-e_m)||^2 = ||e_n||^2 + ||-e_m||^2 = 1^2 + 1^2 = 2 $$ and thus $||e_n - e_m|| \rightarrow \sqrt{2}$ for $n,m \rightarrow \infty$, i.e. $e_n$ is not cauchy. However, I have not only showed that the sequence $e_n$ is not cauchy. But I am not sure whether this implies that a subsequence of $e_n$, i.e. $e_{n_i}$ is not cauchy. How do I conclude or go forward?

Also I have to give an example of a Hilbert space where $B$ is not compact. Any suggetions?

Thanks in advance.

You have proved that the distance between any two distinct terms of that sequence is $\sqrt2$. So, this still holds for any subsequence of that sequence, and therefore no subsequence of that sequence is a Cauchy sequence.