if $(X,A)$ has homotopy extension, so does $(X \cup CA,CA)$ I guess I could use the property: The homotopy extension property for $(X,A)$ is equivalent to $X\times \{0\}\cup A\times I\ $ being a retract of $X\times I$.
Then there is a retraction:$$X\times I\rightarrow X\times \{0\}\cup A\times I$$
I need to show$$(X\cup CA)\times I\rightarrow (X\cup CA)\times \{0\}\cup CA\times I$$is  a retraction too. But I don't know what to do next. Thank you for your time

Putting it more generally, if $(X,A)$ has the HEP, and $f:A \to B$ is a map, then $(B\cup_f X,B)$ has the HEP. The proof is analogous to that of Stefan, which is the case $B=CA$. 
You need the method of defining homotopies on adjunction spaces, which again uses the local compactness of $I$.  
