Suppose $A \subset V$. How does a deformation retraction of $V$ onto $A$ induces a deformation retraction of $V/A$ onto $A/A$? Suppose $A \subset V$.
If there is a deformation of $V$ onto $A$, then there exists maps $i: A \hookrightarrow V$ and $r: V \to A$ such that $ri=Id_A$ and $ir \simeq Id_V$.

How does this deformation retraction of $V$ onto $A$ induces a deformation retraction of $V/A$ onto $A/A$?

I think the map $i': A/A \to V/A$ is just $i'(A)=i(A)=A$ and the map $r': V/A \to A/A$ is $r'([v])=[r(v)]=A$. Is this correct? These maps are well-defined since $A/A$ is just the one point space $\{A\}$.
We have maps $i': A/A \hookrightarrow V/A$ and $r': V/A \to A/A$ such that $r'i'=Id_A$. But how do we show $i'r': V/A \to V/A$ is homotopic to the identity of $V/A$?
 A: Wow I would  be so  depressed if this were false.
One can take a homopy $F:X \times I \to X$ that is a deformation retract. Compose this with the quotient map $q:X \to X/A$, and define $G:=q \circ F$. At each step $t$, there is an induced map 
$\tilde{G}_t:X/A \to X/A$ by the universal property of a quotient map.
Of course at $t=1$ this is a point $a$, and for the intermediate $t$, this point must be fixed since $F_t(A) \subset A$ at all time steps, so $G_t(A)=a$, and so $\tilde{G}_t(a)=a$ for all $t$. In other words, this is a deformation retract.
A: This answer contains essentially everything you need, but it skips the hardest part. I reproduce and complete the argument given there.
You want to prove the following statement:

A deformation retract 
  $$F:V \!\times \! I \to V$$
  of $V$ onto $A$ induces a deformation retract
  $$ \tilde{F} : V/A \times I \to V/A $$
  of $V/A$ onto $A/A$.

Combining the map $F$ with the quotient map $q : V \to V/A$ we get the map 
$$F' : V\!\times\! I \to V/A$$ 
Consider the partition of $V \!\times\! I$ given by the equivalence relation
$$ (x,t) \sim (y,s) \Leftrightarrow t = s \ \text{ and } \ x, y \in A$$
It is easy to see that $F'$ is constant on each element of that partition. This implies that $F'$ can be factorized through $(V \times I)/ \!\! \sim$, i.e. there is a map 
$F'' : (V \! \times\! I)/ \!\! \sim \, \to V/A \ $ 
such that $ \ F'' \circ q = F'$. 
The only thing that is left to show is that 
$(V \!\times\! I)/ \!\! \sim$ 
is homeomorphic to 
$(V/A) \!\times\! I$. 
Combining this homeomorphism with $F''$ we get the desired
map $\tilde{F}$.
Here is one way to do that:


*

*Observe that $V/A$ is the coequalizer of $\, p_1, p_2 : A \!\times\! A \to V$ given by
$\, p_1(a, a') = a \, $ and $\, p_2(a,a') = a'$. 

*Since $I$ is locally compact, the functor $\, -\!\times\! I : \textbf{Top} \to \textbf{Top} \,$ is left adjoint to $\, (-)^I : \textbf{Top} \to \textbf{Top} \,$. In particular, the functor $\, -\!\times\! I$ preserves all existing colimits. 
Therefore $(V/A) \!\times\! I$ is a coequalizer on the maps 
$$\, p_1 \!\! \times \! 1, \, p_2 \!\! \times \! 1: A \!\times\! A \!\times\! I \to V\times I$$
Since $(V \!\times\! I)/ \!\! \sim$ is another coequalizer on the same maps, we get the desired homeomorphism
$$(V/A) \!\times\! I \cong (V \!\times\! I)/ \!\! \sim $$
Note: Compactness of $I$ is used in the following form: the natural bijection
$$ \text{Hom}_\textbf{Top}(X \!\times\! Y, Z) \to \text{Hom}_\textbf{Top}(X, Z^Y),$$
where $Z^Y$ is endowed with the compact-open toplogy,
is given by $f \mapsto (f' : x \mapsto f(x,-)) $. If $f$ is continuous, then so is $f'$. However, the converse does not hold in general. Assuming that $Y$ is locally compact solves the problem: continuity of $f'$ implies the continuity of $f$.
