The essential reason for preferring reduced homology (as experts do)
is that the suspension axiom holds in all degrees, as it must when
one generalizes from spaces to spectra and studies generalized
homology theories. Also, when using reduced homology, one need not
explicitly use pairs of spaces since $H_*(X,A)$ is the reduced
homology of the cofiber $Ci$ of the inclusion $i\colon A\to X$.
The Eilenberg-Steenrod axioms for homology theories have a variant
version for reduced theories, and the reduced and unreduced theories
determine each other. (See for example my book ``A concise course
in algebraic topology'').
Suppose that $i:A\subseteq X$ is a subspace inclusion which is both a cofibration and a homotopy equivalence. We place no other assumptions on $X$ or $A$. The assumptions grant a map $r':X\rightarrow A$ such that $r'i\simeq id_A$ and $ir'\simeq id_X$. Using the homotopy extension property we obtain a map $r:X\rightarrow A$ such that $r\simeq r'$ and $ri=id_A$. Necessarily $ir\simeq id_X$.
$A$ is a deformation retraction of $X$.
Now fix a homotopy $F:id_X\simeq ir$. Define
$$G:X\times\{0,1\}\times I\cup A\times I\times I\rightarrow X$$
by putting
$$G(x,0,t)=x,\qquad G(a,s,t)=F(a,(1-t)s),\qquad G(x,1,t)=F(r(x),1-t).$$
This is well-defined and continuous. We need the following.
Lemma: Let $X$ be a space and $A\subseteq X$ a cofibration. Then $X\times\{0,1\}\cup A\times I\subseteq X\times I$ is a cofibration. $\square$
The proof makes use of the fact that there is a self-homeomorphism of $I^2$ which maps $I\times\{0,1\}\cup\{0\}\times I$ onto $\{0\}\times I$. I'll leave details to your textbook.
Above we have defined $G$. Notice that its restriction to $(X\times\{0,1\}\cup A\times I)\times\{0\}$ is just the map $F$. Because of the lemma there exists a homotopy $\widetilde G:X\times I\times I\rightarrow X$ with $\widetilde G(x,s,0)=F(x,s)$ and $H|_{X\times\{0,1\}\times I\cup A\times I\times I}=G$. Put
$$H_t(x)=\widetilde G(x,t,1).$$
This yields $H_0=id_X$, $H_1=ir$ and $H_t|_A=i$. Thus we have our conclusion.
A subspace inclusion $A\subset X$ which is both a cofibration and a homotopy equivalence is a strong deformation retract.
Of course if $X$ is contractible, then the inclusion $x_0\hookrightarrow X$ of any point is a homotopy equivalence. If this inclusion is a cofibration, then the outcome above applies. The Hausdorff and normality assumptions are not necessary.
Best Answer
Thanks to @Danu for the reference. I will give an outline of the proof in Hatcher:
Claim: A map $f : X \to Y$ induces a homomorphism $f_*: H_n(X) \to H_n(Y)$.
Proof: First, define an induced homomorphism $f_\# : C_n(X) \to C_n(Y)$ by $(\sigma : \Delta^n \to X) \mapsto (f \circ \sigma : \Delta^n \to Y)$.
Secondly, we claim that $f_* : H_n(X) \to H_n(Y) : [a] \to [f_\#(a)]$ is a homomorphism. We need that $f_\#(a) \in \ker \partial_n$ whenever $a \in \ker \partial_n$ (so the map makes sense) and $f_\#(b) \in \text{Im }\partial_n$ whenever $b \in \text{Im }\partial_n$ (so the map is well defined, independent of the choice of representative $a$). ($\partial$ is the differential function $\partial_n : C_n(X) \to C_{n-1}(X)$ or $\partial_n : C_n(Y) \to C_{n-1}(Y)$.)
We can check $f_\# \partial = \partial f_\#$ by explicit computation and this gives us what we need.
We have two easy results:
1) $(f \circ g)_* = f_* \circ g_*$ and
2) id$_{*X} = \text{id}_{H_n(X)}$ (the induced map of the identity on $X$ is the identity on $H_n(X)$).
Proof of 1): We have $(f \circ g)_\# = f_\# \circ g_\#$ (by associativity of $\circ$) and therefore $(f \circ g)_* = f_* \circ g_*$.
Proof of 2): By definition id$_{\#X}$ is id$_{C_n(X)}$ and so id$_{*X} = \text{id}_{H_n(X)}$.
Theorem: If two maps $f,g : X \to Y$ are homotopic, then they induce the same homomorphism $f_* = g_* : H_n(X) \to H_n(Y)$.
The proof of this is the meat of the problem and is given in Hatcher (p111-113).
Corollary: The maps $f_* : H_n(X) \to H_n(Y)$ induced by a homotopy equivalence $f: X \to Y$ are isomorphisms for all $n$. Proof: If $f$ is a homotopy equivalence then there is a map $g$ such that $f \circ g$ is homotopy equivalent to id$_Y$ and $g \circ f$ is homotopy equivalent to id$_X$.
Then, by the theorem, $(g \circ f)_* = id_{*X}$ and $(f \circ g)_* = id_{*Y}$. Using the two easy results, we get $g_* \circ f_* = id_{H_n(X)}$ and $f_* \circ g_* = id_{H_n(Y)}$.
Note that a deformation retraction is a special case of a homotopy equivalence so I've now also shown that homologies are invariant under deformation retraction.