I have seen some answers online regarding this but I wanted to make sure that my proof was correct. I believe this only works if we assume that we have a finite sheeted covering
Now we wish to see why a covering space of a CW complex is also a CW complex. For the $0$-cells of $p:E\rightarrow B$ take them to be the pre-image under $p$ of the $0$-cells of $B$. For the $1$-cells notice that a $1$-cell of $B$ is a map $\phi: I\rightarrow B$ and by the unique lifting property, for each $x\in p^{-1}(b)$ we can find a a lift $\tilde \phi $ starting at $x$. Now for the characteristic maps of dimension $n\geq 2$ we can just use the unique sphere filling property. Now the tricky part is to check that the topology on $E$ is the weak topology. If we take an open set $A$ in the initial topology of $E$ we will have by continuity that it's pre-image by any characteristic map will be open, and so this tells us that $A$ is open in the weak topology.Now take a set $A$ open in the weak topology and we wish to show that it's open.For this we need only to check that $A\cap p^{-1}(U)$ is open for every evenly covered neighborhood $U$. We know that $p^{-1}(U)\cong \cup_i U_i$ with each $U_i$ projecting homeomorphically to $U$. Notice that the set $p(A)\cap U$ is open in $B$ since $A$ is open in the weak topology of $E$ and by definition of a lift $\tilde \phi$ we have that $\phi=p\circ \tilde \phi$ (This is where we use the hypothesis of finite sheets since $p^{-1}(A\cap U)=A\cap \cup_i U_i$). Now notice that $A\cap U_i=p|_{U_i}^{-1}(p(A)\cap U)$. And so $A\cap U_i$ will be open in $U_i$ and since $U_i$ is open it will be open in $X$. Since this is true for any $U_i$ we get that $A\cap p^{-1}(U)$ is open for any evenly covered neighborhood $U$. And so we obtain that the topologies coincide.
Any comment is appreciated. Thanks in advance.