On page $16$ of Hatcher's, following proof is given:
but I can't see the induced deformation retract's existence.
My attempt is to use "passing to the quotient", but somehow at the end I can't guarantee the final step of my argument:
$\require{AMScd}$ \begin{CD} {X_0\sqcup (X_1 \times I)} @>{\text{fix }X_0,\, D = D_2\circ D_1}>> X_0\sqcup (X_1 \times I)\\ @VVV @VVV\\ {X_0\sqcup_{F} (X_1 \times I)} @>{\exists ?}>> {X_0\sqcup_{F} (X_1 \times I)} \end{CD}
Where $D_1$ is the deformation retraction mentioned in the picture onto $X_1\times\{0\} \cup A\times I$, and $D_2$ deforms $X_1\times\{0\} \cup A\times I$ onto $X_1\times \{0\}$ in the natural way.
To apply passing to the quotient argument, I must show (necessarily, but not enough) that whenever $\pi((a,t)) = \pi((a',t'))$ for $a,a \in A$,we have $D((a,t)) = D((a',t'))$. Now condition using $\pi$ is equivalent to $F(a,t) = F(a',t')$. Condition using $D$ is equivalent to $f(a) = f(a')$. But we only have $f_t(a) = f_{t'}(a')$. How to resolve this?
In general, I just can't see how to induce the deformation retraction, so if this is not the way to go, that fact constitutes an answer too.
$\require{AMScd}$I don't quite buy your diagrams because you forget you must in fact have a map $(X_0\sqcup_F(X_1\times I))\times I\to X_0\sqcup_F(X\times I)$ to speak of a deformation retraction.
Let $G:X_1\times I\times I\to X_1\times I$ be the deformation retraction onto $X_1\times\{0\}\cup A\times I$.
Luckily, $-\times I$ is cocontinuous ($I$ is LCH) which is a miracle that makes a lot of stuff in topology work. Thus $(X_0\sqcup_F(X_1\times I))\times I$ is the pushout: $$\begin{CD}A\times I\times I@>>>X_1\times I\times I\\@VF\times1VV@VVV\\X_0\times I@>>>(X_0\sqcup_F(X_1\times I))\times I\end{CD}$$To get a map into $X_0\sqcup_F(X_1\times I)$ again we specifiy the following components; the projection $X_0\times I\to X_0\to X_0\sqcup_F(X_1\times I)$ and $X_1\times I\times I\overset{G}{\to}X_1\times I\to X_0\sqcup_F(X_1\times I)$. As $G$ is identity on $A\times I$ this truly, by pushout properties, induces something $H:(X_0\sqcup_F(X_1\times I))\times I\to X_0\sqcup_F(X_1\times I)$ which is necessarily (using uniqueness of components for pushouts) equal to the identity at time zero and, at time $1$ the image of $H$ is contained in the images of $X_0,X_1\times\{0\}$ and $A\times I$, but $A\times I$'s image is wholly contained in $X_0$ and $X_1\times\{0\}$'s image wholly contained in $X_0\sqcup_fX_1$; we see $H$ at time $1$ has image contained in $X_0\sqcup_f X_1\hookrightarrow X_0\sqcup_F(X_1\times I)$. Moreover, $H$ restricted to this subspace, at any time, yields - as $G$ is identity on $X_1\times\{0\}$ - the identity on both the $X_0$ and $X_1$ components i.e. $H$ is the identity on this subspace, as required.