Completeness proof? First of all, this is not a question about a specific problem, but more about a general technique. When I face a problem such as "show that a metric space $(M,d)$ is complete", the first thing I do is to say: if a metric space is complete, then every Cauchy sequence $(x_n)$ where $x_n\in M$ for all $n$ converges in $M$. From there, I am clueless as to how to proceed: are we allowed to assume that $(x_n)$ at least converges to something that may or may not be in $M$?
 A: It is not exactly a standard method, but there is some grain of truth in what you write. Every metric space can be seen as a sub-space of a complete metric space. Mre precisely, every metric space is isometrically equivalent to a dense subset of a complete metric space and this larger metric space is unique up to isometries.
See the wikipedia page on metric completion.
A: No -- that's not quite it. I think that you are contemplating the fact that we can prove $\mathbb{Q}$ is not complete by finding a sequence in $\mathbb{Q}$ which we know has a limit in $\mathbb{R}$ but not in $\mathbb{Q}$.  But we can't quite work that strongly by analogy here, because we don't know that $M$ lives in some "larger" space where it has a limit.
Instead, you need to start from the assumption only that $(x_n)$ is a Cauchy sequence in $M$ (that is: given $\epsilon>0$, there is $N\in\mathbb{N}$ such that $\lvert x_n-x_m\rvert<\epsilon$ whenever $n,m\geq N$), and show that the sequence must have a limit.
If you have a more specific question in mind, let me know and I'll be happy to try to give you a hint on how to proceed.
A: In practice, you often construct your limit object. It can be by using the completeness of a well-known space, generally $\mathbb{R}$ itsel. You probably know the following examples but the principle is very general. 
Consider $C_b$ the space of continuous and bounded functions from $\mathbb{R}$ to itself, with the norm of uniform convergence $\lVert \cdot \lVert_\infty$. Given a Cauchy sequence $f_n$, you know then that $f_n(x)$ is a real Cauchy sequence for all $x$ and it allows you to defined $f(x)$ as the limit (using that $\mathbb{R}$ is complete) of this sequence. Thus, you have define a function $f$ and then can show that your Cauchy sequence converges to that limit. 
Another example that doesn't use the completeness of $\mathbb{R}$. To show that the space $L^p$ of measurable functions with finite $L^p$-norm is complete, you can proceed the following way. Given a Cauchy sequence for the $L^p$ norm, you show that a subsequence converges almost surely, which defines clearly your limit object, and then show that the whole sequence converges.
Very often, you will have to follow that plan: construct your limit, and show that your
A: Classic three-steps proof once you take a Cauchy sequence in your space:  
(i) find a candidate for your limit (limit something that sounds the right possible limit for the third step),  
(ii) show it belongs to the space (definition of being in the space),  
(iii) show your sequence actually converges to it (definition of cenvergence).
