I'm studying Fourier analysis, and measure theory is too advanced a topic for me, so please give me a rather intuitive explanation.
I cannot understand how this idea applies to a subset of rational numbers.
I was given this description of that fact: 
As I understand it, for compact sets we can find N open covers that will cover all the elements of the set, and shrinking them as we want, but keeping the number of covers, we will still cover all the elements, is this implied here?
If it so, how can we shrink open covers and keeping the number of them to cover the set $\{0\} \cup \left\{2^{-n} :n\in ℕ\right\}$
Measure is in some ways a generalization of the natural and intuitive mathematical notions of "length", "volume" et cetera. Thankfully, we don't actually require that robust of a theory of measure to reap the benefits of what it is to be "measure zero".
To start, it may to be useful to glean some intuition as to what we are representing here. Take for instance the set $\mathbb{Q} \cap [0,1]$. Though it is dense on this interval, there is an intuition that this set is in some ways "small", i.e. it is a countable subset of an uncountable interval. To match this intuition, we seek to develop some theory. This theory starts with some naive notions of volume that are easily computable, e.g. the length of an interval, or in general the volume of a $n$-dimensional rectangle. The idea here is that if we can cover some set with such closed rectangles, and make their volume arbitrarily small, it demonstrates that the covered set is in some way small.
In particular, keeping with the prior example, if we regard $\mathbb{Q} \cap [0,1]$ as the countable union of its composite rational points, say $\bigcup_{i = 1}^{\infty} q_i$, we can for each rational point and arbitrary positive $\epsilon$ consider the enveloping interval $[q_i - \frac{\epsilon}{2^{i+2}}, q_i + \frac{\epsilon}{2^{i+2}}]$ with length $\frac{\epsilon}{2^{i+1}}$ (we can do this as the set is countable). It follows that we have:
$$\mathbb{Q} \cap [0,1] = \bigcup_{i = 1}^{\infty} q_i \subset \bigcup_{i = 1}^{\infty} [q_i - \frac{\epsilon}{2^{i+2}}, q_i + \frac{\epsilon}{2^{i+2}}]$$ Where we additionally have the sum of lengths: $$ \sum_{i =1}^{\infty} \frac{\epsilon}{2^{i+1}} = \frac{\epsilon}{2} < \epsilon $$ And thus $\mathbb{Q} \cap [0,1]$ is suitably small, and we regard this with the title "measure zero".
With regard to how compactness comes into play here (in general, as of course the provided example is not compact), it is relatively straightforward. If a set is already measure zero, we know that we have some cover of intervals/retangles with total length less than say $\frac{\epsilon}{2}$. We can then cover those rectangles with open rectangles of arbitrarily little additional volume (say that the total $< \epsilon$), which of course still cover the original set. As the set is compact (and the covering by these new rectangles is an open cover) then, we are guaranteed a finite subcover of open rectangles that is still subject to this total volume constraint. Inclusion of the boundaries of each of these rectangles then yields a finite covering by closed rectangles of the same total volume. This criteria is referred to in literature some times as "Jordan content zero".
This, in totality, gives you a nonconstructive way to answer your question (that is, it just implies such a finite cover exists fulfilling arbitrarily small total length, as it is quite simple to show your provided set is at least measure zero with a nearly identical argument to the provided one). However, you now have the tools to address to your final question in a more satisfying way. In particular, I may note as a hint of sorts that $2^{-n}$ goes to zero rather quickly. Could you perhaps find a system to cover the points very close to $0$ and then constrain the outliers in suitably small intervals? In particular, you may note that even an arbitrarily small interval about $0$ only fails to cover a finite amount of points in this set.