[proofplan]
We decompose $S^1$ into two open arcs $A$ and $B$, each contractible (homeomorphic to $\mathbb{R}$), overlapping in two disjoint arcs near two chosen points $p$ and $q$. The Mayer--Vietoris sequence then connects $H_*(S^1)$ to the homology of $A$, $B$, and $A \cap B$. Since $H_i(A) = H_i(B) = 0$ for $i \geq 1$, the sequence gives an isomorphism $H_1(S^1) \cong \ker(i_{A*} \oplus i_{B*})$ in $H_0(A \cap B) \cong \mathbb{Z}^2$. We compute this kernel to be $\mathbb{Z}$, generated by $[p] - [q]$. The higher homology groups vanish, and $H_0(S^1) \cong \mathbb{Z}$ by path-connectedness.
[/proofplan]
[step:Set up the open cover and identify the homology of each piece]
Let $p = (1, 0)$ and $q = (-1, 0)$ be two antipodal points on $S^1$. Define $A$ and $B$ to be open arcs that slightly overlap near $p$ and $q$:
\begin{align*}
A &:= \{(\cos\theta, \sin\theta) : -\varepsilon < \theta < \pi + \varepsilon\}, \\
B &:= \{(\cos\theta, \sin\theta) : \pi - \varepsilon < \theta < 2\pi + \varepsilon\},
\end{align*}
for some small $\varepsilon > 0$. Then $A$ and $B$ are open arcs in $S^1$, each homeomorphic to an open interval in $\mathbb{R}$, and $A \cup B = S^1$.
Since each arc is homeomorphic to $\mathbb{R}$ (which is contractible):
\begin{align*}
H_i(A) = H_i(B) = \begin{cases} \mathbb{Z} & i = 0 \\ 0 & i \geq 1. \end{cases}
\end{align*}
The intersection $A \cap B$ consists of two small open arcs: one near $p$ and one near $q$. Each arc is contractible, and the two arcs are disjoint, so $A \cap B$ has two path components:
\begin{align*}
H_i(A \cap B) = \begin{cases} \mathbb{Z} \oplus \mathbb{Z} & i = 0 \\ 0 & i \geq 1, \end{cases}
\end{align*}
with $H_0(A \cap B)$ generated by $[p]$ and $[q]$.
[/step]
[step:Write down the Mayer--Vietoris sequence and show $H_i(S^1) = 0$ for $i \geq 2$]
The Mayer--Vietoris sequence reads:
\begin{align*}
\cdots \to H_i(A) \oplus H_i(B) \to H_i(S^1) \xrightarrow{\partial} H_{i-1}(A \cap B) \to H_{i-1}(A) \oplus H_{i-1}(B) \to \cdots
\end{align*}
For $i \geq 2$, we have $H_i(A) = H_i(B) = 0$ and $H_{i-1}(A \cap B) = 0$ (since $i - 1 \geq 1$). The sequence gives
\begin{align*}
0 \to H_i(S^1) \to 0,
\end{align*}
so $H_i(S^1) = 0$ for all $i \geq 2$.
[/step]
[step:Compute $H_1(S^1) \cong \mathbb{Z}$ from the Mayer--Vietoris exact sequence in low degrees]
The relevant low-degree portion of the Mayer--Vietoris sequence is:
\begin{align*}
0 = H_1(A) \oplus H_1(B) \to H_1(S^1) \xrightarrow{\partial} H_0(A \cap B) \xrightarrow{i_{A*} \oplus i_{B*}} H_0(A) \oplus H_0(B) \to H_0(S^1) \to 0.
\end{align*}
Substituting the known groups:
\begin{align*}
0 \to H_1(S^1) \xrightarrow{\partial} \mathbb{Z} \oplus \mathbb{Z} \xrightarrow{i_{A*} \oplus i_{B*}} \mathbb{Z} \oplus \mathbb{Z} \to H_0(S^1) \to 0.
\end{align*}
**Computing $i_{A*} \oplus i_{B*}$.** The inclusion $i_A: A \cap B \hookrightarrow A$ sends both points $p$ and $q$ into the path-connected space $A$. Since $A$ is path-connected, $i_{A*}([p]) = i_{A*}([q]) = [\text{generator of } H_0(A)]$. Identifying $H_0(A) \cong \mathbb{Z}$ via this generator, $i_{A*}$ sends both $[p]$ and $[q]$ to $1$. Similarly, $i_{B*}$ sends both $[p]$ and $[q]$ to $1$ in $H_0(B) \cong \mathbb{Z}$.
Under the identifications $H_0(A \cap B) \cong \mathbb{Z}^2$ (generated by $[p]$ and $[q]$) and $H_0(A) \oplus H_0(B) \cong \mathbb{Z}^2$, the map $i_{A*} \oplus i_{B*}$ is
\begin{align*}
(a, b) \mapsto (a + b, a + b),
\end{align*}
which has kernel $\{(a, -a) : a \in \mathbb{Z}\} \cong \mathbb{Z}$, generated by $(1, -1)$.
**Extracting $H_1(S^1)$.** By exactness at $H_0(A \cap B)$:
\begin{align*}
\operatorname{im}(\partial) = \ker(i_{A*} \oplus i_{B*}) \cong \mathbb{Z}.
\end{align*}
By exactness at $H_1(S^1)$: $\ker(\partial) = \operatorname{im}(H_1(A) \oplus H_1(B) \to H_1(S^1)) = \operatorname{im}(0) = 0$, so $\partial$ is injective. An injective map with image $\mathbb{Z}$ has domain $\mathbb{Z}$:
\begin{align*}
H_1(S^1) \cong \operatorname{im}(\partial) \cong \mathbb{Z}.
\end{align*}
[guided]
The key computation is determining the kernel of $i_{A*} \oplus i_{B*}: H_0(A \cap B) \to H_0(A) \oplus H_0(B)$.
The intersection $A \cap B$ has two path components, giving $H_0(A \cap B) \cong \mathbb{Z}[p] \oplus \mathbb{Z}[q]$. Each inclusion $i_A, i_B$ maps both components into a single path-connected space ($A$ or $B$), so on $H_0$, both generators $[p]$ and $[q]$ are sent to the same generator.
In matrix form with basis $\{[p], [q]\}$ for $H_0(A \cap B)$ and the natural generators for $H_0(A) \oplus H_0(B)$:
\begin{align*}
i_{A*} \oplus i_{B*} = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}: \mathbb{Z}^2 \to \mathbb{Z}^2.
\end{align*}
The kernel of this matrix is $\operatorname{span}_\mathbb{Z}\{(1, -1)\} \cong \mathbb{Z}$.
The connecting homomorphism $\partial: H_1(S^1) \to H_0(A \cap B)$ maps into this kernel. Since $\partial$ is injective (the group to its left in the exact sequence is $0$), $H_1(S^1)$ is isomorphic to a subgroup of $\mathbb{Z}$, which is $\mathbb{Z}$ itself (since $\operatorname{im}(\partial) = \ker(i_{A*} \oplus i_{B*}) \cong \mathbb{Z}$ and injective maps give isomorphisms onto their images).
Geometrically, the generator of $H_1(S^1)$ corresponds to "going around the circle once": the connecting homomorphism sends this cycle to $[p] - [q]$, recording that the cycle enters one component of $A \cap B$ and exits the other.
[/guided]
[/step]
[step:Conclude with $H_0(S^1) \cong \mathbb{Z}$]
Since $S^1$ is path-connected, $H_0(S^1) \cong \mathbb{Z}$.
Alternatively, from the Mayer--Vietoris sequence: by exactness, $H_0(S^1) \cong \operatorname{coker}(i_{A*} \oplus i_{B*})$. The image of $i_{A*} \oplus i_{B*}$ is $\{(a + b, a + b) : a, b \in \mathbb{Z}\} = \{(c, c) : c \in \mathbb{Z}\}$. The quotient $\mathbb{Z}^2 / \{(c, c)\} \cong \mathbb{Z}$ via $(a, b) \mapsto a - b$. This confirms $H_0(S^1) \cong \mathbb{Z}$.
Combining all degrees:
\begin{align*}
H_i(S^1) = \begin{cases} \mathbb{Z} & i = 0, 1 \\ 0 & \text{otherwise.} \end{cases}
\end{align*}
[/step]
