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
 A: 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 $I$.  
A: Note that $(X,A)$ having the HEP implies that $X×\{0\}\cup A×I$ has the coherent topology with respect to the covering $(X×\{0\},A×I)$, that means a function from that space is continuous if its restrictions to $X×\{0\}$ and to $A×I$ are continuous on the respective spaces (so that space is homeomorphic to $M_i$, the mapping cylinder of the inclusion $i:A\to X$). 
Consider the following commutative diagram
$$
\begin{array}
@ X\times I \sqcup CA×I  & \longrightarrow & (X∪CA)×I \\
\downarrow r\sqcup 1_{CA} & & \downarrow r' \\
(X×\{0\}\cup A×I) \sqcup CA×I & \longrightarrow &(X∪CA)×\{0\}∪CA×I
\end{array}
$$
Since $I$ is locally compact, the upper map is a quotient map identifying
$$
\qquad\qquad (a,t)\sim([a,0],t) \qquad\qquad (1)
$$
for any $a\in A$ and $t\in I$. The map $r\sqcup 1_{CA×I}$ sends
$$(x,t)\mapsto r(x,t),\quad ([a,s],t)\mapsto ([a,s],t)
$$
where $r$ is the retraction. Finally, the lower map is defined on 
$X×\{0\}\cup A×I$ by sending 
$$
(x,0) \mapsto (x,0) \in (X\cup CA)×\{0\}, \quad 
(a,t) \mapsto ([a,0],t) \in CA×I
$$
and it is continuous by the remark at the beginning of my post.
Since the identifications (1) are respected by the lower left path, the map $r'$ is induced.
