Relation between Compactness, Closedness and Completness of metric spaces I would like to know as many relations as possible to get a better picture.
I know that if $f$ is continuous and $(X,d)$ is complete, then $f(X)$ is complete $\iff$ closed.
Question:However, are complete sets closed? Since complete sets are when Cauchy sequences converge in those sets, it would make sense for them to be closed (since they contain their limit points)?
Question: I know that compact sets are closed. Take for example $\{0\cup\frac{1}{n}\}$. If I take $x_n=1/n$ I see that $x_n\rightarrow 0$ which means that it's closed. But is it compact? What cover can I take? Will $B(0,r=1)$ will do? If yes, then it's bounded and I could ?possibly? use Heine Borel to show it's compact?
Example: Is $C[0,1]$ compact? Could you provide me with an example of interesting non-compact spaces? It seems I am only working with intervals
 A: Question 1: Any convergent sequence of some metric space is a Cauchy sequence in this metric space. If any Cauchy sequence in this space has a limit in this space (i.e. if the metric space is complete), it means in particular that any convergent sequence has its limit in this space, hence the metric space is closed.
Question 2: I assume you consider $\{0\}\cup\{1/n\mid n\in\mathbb{N}_{0}\}$ as a subset of $\mathbb{R}$ equipped with the euclidean distance (and seen as a topological space for the topology induced by this distance) with the induced topology. In this case, as your subset $\{0\}\cup\{1/n\mid n\in\mathbb{N}_{0}\}$ is bounded and closed, by Heine-Borel you know it is indeed compact.
Question 3: I assume you look for bounded closed subsets of a topological space that are not compact in this topological space (because if we can take unbounded closed subsets, taking $\mathbb{R}$ with the classical topology induced by the euclidean distance, it is non-compact). I do not know if you have knowledge of measure theory and functional analysis, but in these fields arise infinite dimensional normed vector spaces whose unit ball is not compact. As I said in the comments, a normed vector space is finite dimensional if and only if its closed unit ball is compact (the proof is not complex).
For your particular example, with which topology do you equip $C[0,1]$?

For an easier example, take the space $\ell^{2}$ of all sequences of reals $(x_{n})_{n}$ such that 

$$\Vert (x_{n})_{n}\Vert_{\ell^{2}}=\sqrt{\sum_{i=1}^{\infty}\vert x_{i}\vert^{2}}<\infty$$
Such a sequence is said to be "square-summable" and the function:
$$\ell^{2}\to\mathbb{R}^{+}:(x_{n})_{n}\mapsto\Vert (x_{n})_{n}\Vert_{\ell^{2}}=\sqrt{\sum_{i=1}^{\infty}\vert x_{i}\vert^{2}}$$
defines a norm on $\ell^{2}$, which induces a natural structure of metric space (and thus of topological space) with distance $d(x,y)=\Vert x-y\Vert_{\ell^{2}}$ where $x,y$ are two sequences of $\ell^{2}$. 
One can show that $\ell^{2}$ is complete. However, the closed unit ball 
$$\overline{B(0,1)}=\left\{(x_{n})_{n}\in\ell^{2}\mid \Vert (x_{n})_{n}\Vert_{\ell^{2}}\le 1\right\}$$
is not compact (and, in particular, it not anymore holds that being bounded and closed is equivalent to be compact!).
