Let $X$ be a connected topological space, and $C_{finite}$ the category of its finite coverings. Then I claim that the category $C$ of coverings of $X$ can be obtained by $C_{finite}$ taking projective limits and disjoint unions.
Since projective limits and disjoint unions commute the statement is reduced to say that $A:= \varprojlim A_i$, where each $A_i$ is a finite covering, is itself a covering space of $X$. For the proof this is my problem: Fixed a point $x\in X$, and a point $\{y_i \in A_i\} \in A$ of the fiber over $x$, then I must find an open neighborhood $x\in U \subset X$ such that it is homeomorphic to a convenient neighborhood $V$ of $\{y_i\}\in A$.
The passage is problematic because, if the limits is taken over an infinite index set $I$, then I'm not allowed anymore to restrict this neighborhood $U$ in a convenient way for each $A_i$ (Infinite intersection of opens can be not open).
There is a way to bypass this problem? Or my claim is simply wrong?
Probably in the solution it will be helpful (or necessary) to restrict to a base space $X$ which admits a universal covering, i.e. it is locally path connected and semi-locally simply connected.
Edit: I forgot to mention the following fact, which justify my claim. If $X$ admits a universal covering, then it can be obtained as projective limit of finite (normal) coverings of $X$.