If $(X,E)$ is a CW complex then so is $X \times I$.
I have constructed the characteristic maps:
We could give a cell decomposition of $X \times I$ by $E'' := \{ e \times a^0, e \times b^0 , e \times c^1 , e \in E \}$ where $I = a^0 \cup b^0 \cup c^1$ is a CW decomposition of $I$. For each $n$-cell, of the first two forms, we use the original characteristic map, $$ \Phi_{e\times j} := i_j \circ \Phi_e, \quad j = a^0,b^0$$ For the third type of cells, consider the map $g_n : D^{n+1} \rightarrow D^n \times I$, $x \mapsto |x|f^{-1}\Big(\frac{x}{|x|}\Big)$, where $f: D^n \times I \rightarrow S^{n}$ is restriction of $\frac{x}{|x|}$ map. So, $$\Phi_{e \times c^1} := (\Phi_e \times id_I) g_n$$
yields characteristic maps for $e \times c^1$.
It suffices to show that $X \times I$ has the weak topology wrt cells in $E''$. How does one prove this?
This is nontrivial and uses the (local) compactness of $I$. You can prove it directly by using the tube lemma to build up open sets in the product topology, but here is a cleaner proof using a little theory of function spaces. The key fact we use is that if $X$, $Y$, and $Z$ are spaces with $Y$ locally compact, then a map $X\times Y\to Z$ is continuous iff the induced map $X\to Z^Y$ is continuous, where $Z^Y$ is the space of all continuous maps $Y\to Z$ with the compact-open topology.
Now to show $X\times I$ has the weak topology, it suffices to show it has the weak topology with respect to the maps $\Phi_e\times id_I:D^n\times I\to X\times I$ for each cell $e$ of $X$. (We may consider $\Phi_e\times id_I$ instead of $\Phi_{e\times c^1}$ since the map $g_n$ is a homeomorphism, and this also covers cells of the form $e\times a^0$ or $e\times b^0$ since their characteristic maps are just the restrictions of $\Phi_e\times id_I$ to the endpoints of $I$.)
So we wish to show that for any space $Z$ a map $f:X\times I\to Z$ is continuous iff $f(\Phi_e\times id_I):D^n\times I\to Z$ is continuous for each $e$. Since $I$ is locally compact, $f$ is continuous iff the induced map $\hat{f}:X\to Z^I$ is continuous. Since $X$ has the weak topology, $\hat{f}$ is continuous iff $\hat{f}\Phi_e$ is continuous for each $e$. But $\hat{f}\Phi_e$ is just the map $D^n\to Z^I$ induced by $f(\Phi_e\times id_I):D^n\times I\to Z$, so $\hat{f}\Phi_e$ is continous iff $f(\Phi_e\times id_I)$ is continuous. Thus $f$ is continuous iff each $f(\Phi_e\times id_I)$ is continuous, as desired.
To convert this into an explicit proof in terms of open sets, you can just take $Z=\{0,1\}$ with the topology that $\{1\}$ is open but $\{0\}$ is not, so that continuous maps to $Z$ are the same as characteristic functions of open sets. When you chase through what the argument above gives concretely in that case, you get the following:
Let $U\subseteq X\times I$ be open in the weak topology and let $(x,t)\in U$. Then there is some closed neighborhood $J\subseteq I$ of $t$ such that $\{x\}\times J\subseteq U$. Now let $V$ be the set of $y\in X$ such that $\{y\}\times J\subseteq U$. For each cell $e$ of $X$, $(\Phi_e\times id_I)^{-1}(U)$ is open in $D^n\times I$. By the tube lemma (since $J$ is compact), $\Phi_e^{-1}(V)$ is open in $D^n$, since it is just the set of $y\in D^n$ such that $\{y\}\times J\subseteq (\Phi_e\times id_I)^{-1}(U)$. Since $X$ has the weak topology, this means $V$ is open in $X$. But now $V\times int(J)$ is open in the product topology on $X\times I$ and $(x,t)\in V\times int(J)\subseteq U$. Since $(x,t)\in U$ was arbitrary, this shows $U$ is open in the product topology.