How can I show that the composition of two coverings is also a covering? I'm trying to prove the following:

Let $\varpi ' : X'' \to X'$ and $\varpi : X' \to X$ be two coverings
  and let $X$ be a locally simply connected space. Prove that $\varpi \circ \varpi ' : X'' \to X$ is also a covering.

I am completely stuck at this, I have no idea about how to use the locally simply connected hypotesis about $X$. How should I proceed?
Any hint would be greatly appreciated.
 A: Firts assume that $X$ is simply connected. Then $X'= X\times D$, $D$ a space with the discrete topology. Therefore $X"= X\times D'\times D$ for another discrete space $D'$, and the result follows. The general case follows from the  hypothesis : let $x\in X$ and $U$ a simply connected neigbourhood of $X$, the previous argument applied to $U\subset X$ $\omega^{-1} U \subset X'$ and $\omega "^{-1}(U)\subset X"$ prove that the result.
A: I got the solution to this problem.
$X'$ and $X''$ are locally simply connected because of the definition of covering. Also, we can consider $U \subset X$ simply connected (and thus a trivializing neighbourhood). Then , the preimage of $U$ is:
$$\varpi^{-1}(U) = \sqcup_{i} V_i$$
Where every $V_i$ is homeomorphic to $U$.  In the same way is easy to see that:
$$\varpi ' ^{-1}(V_i) = \sqcup_{j} W_{ij} $$
Then, the problem can be solved because $U$ is a trivializing neighbourhood (here is where I had to use the hypotesis of $X$ being locally simply connected). Then the preimage of $U$ via the composition $\varpi \circ \varpi'$ is:
$$(\varpi \circ \varpi')^{-1}(U) = \sqcup_i \sqcup_j W_{ij}$$
And this works because the composition of homeomorphisms is also an homeomorphism.
