The universal cover covers any connected cover The Wikipedia page about covering spaces asserts that: 

The universal cover (of the space $X$) covers any connected cover (of the space $X$).

I'm not able to find a proof of this result. Any idea how to find it or how to prove it?
Here the definition of universal cover is just that of a covering being simply connected. Moreover, a part from connectedness of the other cover there is no other hypothesis.
 A: Sofie Verbeek's answer is perfect, but let me go back to the definition of a universal covering space. There are two different approaches: Some (perhaps most) authors define a universal covering $p : E \to X$ of a connected space $X$ to be one such that $E$ is simply connected, other authors define it by the property that it covers any covering $p' : E' \to X$ with a connected $E'$.
In my opinion the second definition is the better one because it is explains the name. If you accept this point of view, then it is a theorem that a simply connected covering of a connected and locally path connected $X$ is a universal covering.
A nice reference is Chapter 2 Section 5 of
Spanier, Edwin H. Algebraic topology. Vol. 55. No. 1. Springer Science & Business Media, 1989.
It contains a number of interesting results and examples (e.g. of a non-simply connected universal covering space of a of a connected and locally path connected space).
A: Hint. Covering spaces satisfy the path lifting property. That is, given a path in your base space $X$ from a point $x$ to a point $y$, together with a given lift of $x$ in your covering space $E$, there is a unique lift of the path to the covering space, starting at the lift of $x$. Prove that if your covering space $E$ is simply connected, the end-point of the lifted path does not depend on your initial choice of path from $x$ to $y$. It is at this point that the property of being simply connected is crucial.
Next, use this special property to manually define a map from the simply connected covering $E$ to any other covering space $E'$, and prove that the map $E \to E'$ that you constructed is a covering of $E'$ by $E$.
