Compactness in normed vector spaces.

Yes.

The equivalence of "compact" and "closed and bounded" holds only in finite-dimensional spaces.

In a general, infinite-dimensional normed space (e.g., over the real scalars) complete in its norm (i.e., an infinite-dimensional Banach space), for a set to be compact, it is not enough that it be closed and bounded. It must be closed and totally bounded.

Let $c_{0}$ denote the infinite-dimensional normed vector space consisting of sequences that converge to $0$. The norm on $c_{0}$ is given by:

$$ \|(a_{1},a_{1},a_{3},\ldots)\|:=\sup_{k\in\mathbb{N}}|a_{k}|. $$

Let $A$ denote the set of all vectors in $c_{0}$ of norm at most one. Clearly $A$ is bounded and it is easy to show that $A$ is closed. To see that $A$ is not compact, let, for each $x\in A$, $B_{1/2}(x):=\{y\in c_{0}:\|x-y\|<1/2\}$. Then, $\{B_{1/2}(x):x\in A\}$ is an open cover of $A$. To see that it has no finite subcover, let, for each $k\in\mathbb{N}$, $e_{k}$ denote the sequence $(0,\ldots,0,1,0,\ldots)$, where the $1$ is in position $k$. Since $\|e_{k}-e_{\ell}\|=1$ if $k\not=\ell$, it follows that no two such vectors can belong to the same $B_{1/2}(x)$ for any $x$.

Note that in general, a set in a metric space is compact if and only if it is complete and totally bounded.

Example. The unit ball $B$ in Hilbert space $l^2$. It is closed and bounded, but not compact. Let the orthonormal unit vectors be $e_n, n=1,2,3,\dots$. Then sequence $e_n$ is a sequence in $B$, but it has no convergent subsequence. This is because all the distances are $\|e_n-e_m\| = \sqrt{2}$ so in fact no subsequence is even a Cauchy sequence.