Let
$$ l^2 = \left\{ (x_n) : \sum_{n=1}^{\infty} x_n^2 < \infty \right\} $$
equipped with the norm
$$ \| (x_n) \| = \left( \sum_{n=1}^{\infty} x_n^2 \right)^{1/2}. $$
I am wondering whether the following two subsets of $l^2$ are sequentially compact or not.
$\displaystyle \left\{ (x_n) \in l^2 : \sum_{n=1}^{k} x_n^2 \leq 1 \right\}$ where $k \in \mathbb{N}$ is fixed;
$\displaystyle \left\{ (x_n) \in l^2 : \sum_{n=1}^{\infty} x_n^2 \leq 1 \right\}$
For both sets, you can consider the sequence of sequences $e_i$, where $$ (e_i)_j = \begin{cases} 0 & i \neq j \\ 1 & i=j \end{cases} $$
Note that $||e_i|| = 1$ for all $i$, so that $e_i$ belong in both spaces.However, $||e_i-e_j|| = \sqrt 2$, so no subsequence of $e_i$ is Cauchy, hence not convergent. Therefore, both spaces are not sequentially compact.
None is. Consider the canonical basis $\{\delta_n\} $, where $$\delta_n=(\delta_{n,m})_m. $$ Then $\delta_n $ is in both of your sets, and $$\|\delta_k-\delta_j\|=\sqrt2$$ for all $k,j $; this shows that the sequence does not admit a convergent subsequence. Also, your second set is the unit ball. The unit ball of a normed space is compact if and only if the space is finite-dimensional.
$S:= \{(x_n) \in \ell^2 : \sum_{i=0}^k x_n^2 \le 1 \}$ fails to be compact because it is not bounded. Define $a_\alpha$ to be a sequence with all $0$'s except for an $\alpha$ in position $k+1$. We have $a_\alpha \in S$ for all $\alpha \in \ell^2$ but $\|a_\alpha\| \to \infty$ as $\alpha \to \infty$, so $S$ fails to be bounded.
It is a famous result that the unit ball in any infinite dimensional normed vector space fails to be compact, so the second of your sets fails to be compact.
