Let $X$ be a CW complex, $p:E\to X$ a covering map. Then $E$ has an induced CW complex structure defined as follows. If $\Phi:D^n\to X$ is a covering, it lifts to a map $D^n\to E$ (since $D^n$ is simply connected and we can apply the lifting criterion). These give the desired cell decomposition.
However, I don't know how to prove that $E$ has the weak topology induced by the aforementioned cell decomposition. Somehow I must use the fact that $p:E\to X$ is a covering and that $X$ has the weak topology, but I don't know how...
This question has been asked several times on this site, e.g., here, and here. But (surprisingly) none of the answers there give a proof of this...
Thanks in advance!
Suppose $A\subseteq E$ is open in every cell of $E$ (i.e., $\Phi^{-1}(A)$ is open for each cell $\Phi:D^n\to E$). It suffices to show that $A\cap p^{-1}(U)$ is open for any open $U\subseteq X$ which is evenly covered by $p$, since such sets $p^{-1}(U)$ are an open cover of $E$.
Since $U$ is evenly covered by $p$, we may identify $p^{-1}(U)$ with a disjoint union of copies $U_i$ of $U$. Consider each such copy $U_i$ as sitting inside a copy $X_i$ of $X$. Consider a cell $\Phi:D^n\to X_i$ and a connected component $V$ of $\Phi^{-1}(U_i)$. Then there is a cell $\Phi':D_n\to E$ such that $\Phi'|_V=\Phi|_V$ (just let $\Phi'$ be the lift of $\Phi$ to $E$ that agrees with it on $V$). Since $A$ is open in every cell of $E$, we conclude that $V\cap \Phi^{-1}(A\cap U_i)=V\cap\Phi'^{-1}(A\cap U_i)$ is open. Since $V$ was an arbitrary connected component of $\Phi^{-1}(U_i)$, this means $\Phi^{-1}(A\cap U_i)$ is open.
Thus we have shown that $A\cap U_i$ is open in every cell of $X_i$. Since $X_i$ has the weak topology, this means $A\cap U_i$ is open in $X_i$, and thus open in $U_i$. Since $p^{-1}(U)$ is the disjoint union of the $U_i$, this means $A\cap p^{-1}(U)$ is open in $p^{-1}(U)$ and hence in $E$.