Let $E=\{x=(x_n) \in l^2 : |x_k|\leq\frac{1}{\sqrt{k}}, \forall k \geq 1\}$ be a subset of $l^2$. I have two questions:
1. $E$ is bounded ? I don't think so, because $$ |x|^2= \sum^{\infty}_{n=1}|x_n| \leq \sum^{\infty}_{n=1}\frac{1}{n} \to \infty, \forall x \in E $$ Is this right?
2. $E$ is compact?
Consider the sequence of sequences $\{\mathbf x^{(n)}\}_{n\ge 1}$ given by $\mathbf x^{(n)} = \{\mathbf x^{(n)}_k\}_{k\ge 1}$ where $$\mathbf x^{(n)} = \left(1, \frac{1}{\sqrt{2}},\ldots, \frac{1}{\sqrt{n}},0, 0,0,\ldots\right)$$ Clearly, $\mathbf x^{(n)} \in \ell^2$ for each $n\ge 1$, and $\left|\mathbf x^{(n)}_k\right| \le \frac{1}{\sqrt k}$ for every $k\ge 1, n\ge 1$. Furthermore, we have $$\|\mathbf x^{(n)}\|_2^2 = \sum_{k=1}^\infty \left|\mathbf x^{(n)}_k\right|^2 = \sum_{k=1}^n \frac 1 k \xrightarrow{n\to\infty} \infty$$ as the harmonic series $\sum_{k\ge 1} \frac 1 k$ diverges. Therefore, the sequence $\{\mathbf x^{(n)}\}_{n\ge 1} \subset E$ is unbounded, showing that the set $E$ is not bounded.
Remark: Let $(X,d)$ be a metric space, of which $E\subset X$ is a compact subset. Choose any $x_0\in E$. The balls $\{B(x_0,m)\}_{m=1}^\infty$ constitute an open cover of $E$. By compactness, there exists a finite subcover. In particular, there exists $n\in \Bbb N$ such that $E\subset B(x_0,n)$. By the triangle inequality, we have $d(x,y) < 2n$ for any $x,y\in E$. Thus, $E$ is bounded.