Application of Heine–Borel theorem. I need to decide whether this proposition is true or not :"A countable set is always compact".
My first thought is NO. 
As a simple proof I would present a counter example:  since $\mathbb{N}$ is countable if the claim was true it would imply that it's compact. Compactness (by the of Heine–Borel theorem) requires Closeness (that in this case is satisfied) and Boundness (which fails to hold). For this reason the claim is not true. 
Now, I know that in certain spaces the Heine-Borel property does not apply, and that's why I would like to know if in this very generic case (the exercise does not provide any other information) this "proof" is a valid one.
If not, I would like to know what extra conditions should I impose to make it true.
Thanks in advance.
 A: The open-ness of this question really comes down to what set, and what topology on that set the asker has in mind. If both of those pieces of data are left open to interpretation on the part of the answerer, then your answer is perfectly fine. What your answer shows is that countable sets are not always compact, in the most general sense.
The Heine-Borel theorem says that a subset of $\mathbf R$, equipped with its usual topology, is compact if and only if it is closed and bounded. The collection $\mathbf N$ of natural numbers, thought of as a subset of $\mathbf R$, is countable, closed, and unbounded, so it is not compact by Heine-Borel.

If we wanted to modify the question to only consider certain types of topological spaces, that is, sets equipped with certain kinds of topologies, then maybe we can get different answers. For instance, any countable subset (in fact any subset whatsoever) of a set equipped with the indiscrete topology is compact.
A: Miscellany.
(1). In Sierpinski space $S=\{1,2\}$ the sets $\emptyset, S,$ and $\{1\}$ are open but $\{2\}$ is not open. Any finite subset of any space is a compact subset so $\{1\}$ is a compact subset of $S.$ But $\{1\}$ is not closed because its complement $\{2\}$ is not open. This little space is useful for a large number of examples and counter-examples in topology.
(2)."Bounded" is not a topological property, in the following sense: By examining the family of open sets, we can determine whether a space is compact, and (sometimes needing some  rather deep results) determine whether the topology can be generated by a metric.  But if a space $S$ is metrizable and not compact then (not "obviously") there exists an unbounded metric $d$ on S that generates the topology.  But the metric $e(x,y)=\min (1,d(x,y))$ is bounded and generates the same topology. So we cannot always say whether some  subset of some $S$ is bounded or not by just looking at the topology. It may depend on the metric. 
(3). A space $S$ is $T_2$ (a.k.a. a Hausdorff space ) when distinct points in $S$ have disjoint neighborhoods. I.e. if $x,y\in S$ with $x\ne y$  then there are disjoint open sets $U,V$ with $x\in U$ and $y\in V.$ 
Metric spaces are Hausdorff.
A compact subset of a Hausdorff space is necessarily closed.
(Remark on terminology: A neighborhood $T$ of a point $x$ in a space $S$ is some $T\subseteq S$ such that there exists an open set $U$ with $x\in U\subseteq T.$) 
(4). We can say that a subset $V$ of a metrizable space is compact iff for any sequence  $\sigma=(v_n)_{n\in \Bbb N}$ of members of $V,$ there exists $v(\sigma)\in V$ and a  sub-sequence $(v_{n_i})_{i\in \Bbb N}$ of $\sigma,$ such that $0=\lim_{i\to \infty}d(v_{n_i},v(\sigma)) $ for each and every metric $d$ for the space. (Do not overlook the condition that $v(\sigma)\in V.$)
In brief, in a metrizable space, compactness is equivalent to sequential compactness.
(5). Modern functional analysis often treats certain families of functions as sets of points in a (suitably defined) topological space.
Example: 
$C[0,1]$ is the set of continuous $f:[0,1]\to \Bbb R,$ and the sup norm $\|f-g\|=\sup_{x\in [0,1]}|f(x)-g(x)|$ defines a metric $d(f,g)=\|f-g\|$ on $C[0,1].$ In this metric, convergence of a sequence $(f_n)_{n\in \Bbb N}$ to $f$ means the sequence  $(f_n)_{n\in \Bbb N}$ of functions  converges uniformly to the function $f$.
If $V\subset C[0,1] $ and $V$ has non-empty interior  then $V$ is not compact.  Because, if $f\in V$ and $r>0$ and $\{g: \|g-f\|<r\}\subset V,$ let $f_n(x)=f(x)+rx^n/2$  for each  $n\in \Bbb N$ and $x\in [0,1].$ Then no sub-sequence  of $(f_n)_n$ converges uniformly to  $any $ member of $ C[0,1]$ so $V$ is not sequentially compact, so $V$ is not compact.
In particular $\{g: \|g\|\le 1\}$ is closed and bounded but not compact.  (For more on compactness in this space, look up the Arzela-Ascoli Theorem or the general topic of Normed Linear Spaces, a.k.a. Normed Vector Spaces. ) 
