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.
Let us first look at $ H_0 ( \mathbb{R}^n - \{0\} ) \xrightarrow{(i_0)_{\ast} } H_0 ( \mathbb{R}^n ) \xrightarrow{ (j_0)_{\ast} } H_0 ( \mathbb{R}^n, \mathbb{R}^n - \{0\} ) \xrightarrow{ \partial_0 } 0$, where $n \ge 1$. Take any map $\phi : \mathbb R^n \to \mathbb{R}^n - \{0\}$. Since $\mathbb R^n$ is contractible, we have $i_0 \circ \phi \simeq id$ and therefore $(i_0)_{\ast} \circ \phi_* = id$. This implies that $(i_0)_{\ast}$ is a surjection. Therefore $\ker (j_0)_* = \operatorname{im} (i_0)_* = H_0 ( \mathbb{R}^n )$, i.e. $(j_0)_*$ is the zero map. Thus $\ker \partial_0 = \operatorname{im} (j_0)_* = 0$. Since $\partial_0$ is the zero map, we get $$ H_0 ( \mathbb{R}^n, \mathbb{R}^n - \{0\} ) = \ker \partial_0 = 0 .$$
Therefore we have the short exact sequence
$$
0 \xrightarrow{ (j_1)_{\ast} } H_1 ( \mathbb{R}^n, \mathbb{R}^n - \{0\} ) \xrightarrow{ \partial_1 } H_0 ( \mathbb{R}^n - \{0\} ) \xrightarrow{ (i_0)_{\ast} } \mathbb{Z} \xrightarrow{ (j_0)_{\ast} } 0
$$
We have $\ker \partial_1 = \operatorname{im} (j_1)_* = 0$, i.e. $\partial_1$ is injective.
1) $n \ge 2$. We get
$$
0 \xrightarrow{ (j_1)_{\ast} } H_1 ( \mathbb{R}^n, \mathbb{R}^n - \{0\} ) \xrightarrow{ \partial_1 } \mathbb Z \xrightarrow{ (i_0)_{\ast} } \mathbb{Z} \xrightarrow{ (j_0)_{\ast} } 0 .
$$
We know that $(i_0)_*$ is a surjection. This is only possible if $(i_0)_*(1) = \pm 1$, thus $(i_0)_*$ is an isomorphism. We conclude that $ \operatorname{im}\partial_1 = \ker (i_0)_* = 0$. Since $\partial_1$ is injective, this implies
$$ H_1 ( \mathbb{R}^n, \mathbb{R}^n - \{0\} ) = 0 .$$
2) $n = 1$. We get
$$
0 \xrightarrow{ (j_1)_{\ast} } H_1 ( \mathbb{R}, \mathbb{R} - \{0\} ) \xrightarrow{ \partial_1 } \mathbb Z \oplus \mathbb Z \xrightarrow{ (i_0)_{\ast} } \mathbb{Z} \xrightarrow{ (j_0)_{\ast} } 0 .
$$
Since $\partial_1$ is injective, $H_1 ( \mathbb{R}, \mathbb{R} - \{0\} )$ is isomorphic to the subgroup $G = \ker (i_0)_*$ of $\mathbb Z \oplus \mathbb Z$. The latter is a free abelian group with two generators, thus also $G$ is a free abelian group whose number $g$ of generators is $0, 1$ or $2$.
The above short exact sequence splits because there exists a homomorphism $h : \mathbb Z \to \mathbb Z \oplus \mathbb Z$ such that $(i_0)_* \circ h = id$. This can either be deduced form the fact that $\mathbb Z$ is a free abelian group or by taking the above $\phi_*$. It is well-known that this implies $\mathbb Z \oplus \mathbb Z \approx H_1 ( \mathbb{R}, \mathbb{R} - \{0\} ) \oplus \mathbb Z \approx G \oplus \mathbb Z$. An explicit ismorphism is given by $I : H_1 ( \mathbb{R}, \mathbb{R} - \{0\} ) \oplus \mathbb Z \to \mathbb Z \oplus \mathbb Z, I(\xi,x) = \partial_1(\xi) + h(x)$.
$G \oplus \mathbb Z$ is a free abelian group with $g+1$ generators and we must have $g + 1 = 2$ since $\mathbb Z \oplus \mathbb Z$ has two generators. Hence $g = 1$ and
$$H_1 ( \mathbb{R}, \mathbb{R} - \{0\} ) \approx G \approx \mathbb Z .$$
Remark:
The above proof for 2) was purely algebraic / group theoretical. One can alternatively use singular $0$-simplices to show that $(i_0)_* : \mathbb Z \oplus \mathbb Z = H_0((-\infty,0)) \oplus H_0(0,\infty) = H_0(\mathbb{R}- \{0\}) \to H_0(\mathbb R) = \mathbb Z$ has the form $(i_0)_*(a,b) = a + b$. Then $G = \ker (i_0)_* =\{(a,-a) \mid a \in \mathbb Z\}$. This group is isomorphic to $\mathbb Z$.
Best Answer
The idea behind is the following: suppose $X$ is path connected. $H_0(X)$ is the quotient of the zero cycles $Z_0(X)=\{n_0x_0+...+n_xx_p, x_0,...,x_p\in X\}$ by the image $d_1:C_1(Z)\rightarrow Z_0(X)$ where $C_1(X)$ is module of $1$-chains.
Let $x,y\in X$, there exists a path $c:I=[0,1]\rightarrow X$ such that $c(0)=x,c(1)=y$. $c(I)$ is a $1$-chain and $d_1((c(I))=x_1-x_0$, this implies that $Z_0(X)/d_1(C_1(X))$ is generated by the class of $x$.