[proofplan] We use the contracting homotopy directly. For each degree $n$, every $n$-cycle $z \in \ker d_n$ is shown to be the boundary of the element $s_n(z) \in C_{n+1}$. Hence the cycle subgroup equals the boundary subgroup, so the quotient defining $H_n(C_\bullet)$ is the zero group. [/proofplan]

[step:Define the cycle and boundary subgroups in degree $n$] Fix $n \in \mathbb{Z}$. Define the $n$-cycle subgroup $Z_n(C_\bullet) \le C_n$ and the $n$-boundary subgroup $B_n(C_\bullet) \le C_n$ by \begin{align*} Z_n(C_\bullet) &= \ker(d_n: C_n \to C_{n-1}), \\ B_n(C_\bullet) &= \operatorname{im}(d_{n+1}: C_{n+1} \to C_n). \end{align*} Since $d_n \circ d_{n+1} = 0$, every boundary is a cycle, so \begin{align*} B_n(C_\bullet) \subseteq Z_n(C_\bullet). \end{align*} [/step]

[step:Show that every cycle is a boundary using the contracting homotopy]Let $z \in Z_n(C_\bullet)$. Then $z \in C_n$ and $d_n(z) = 0$. Applying the contracting homotopy identity in degree $n$ gives \begin{align*} z &= \operatorname{id}_{C_n}(z) \\ &= d_{n+1}(s_n(z)) + s_{n-1}(d_n(z)) \\ &= d_{n+1}(s_n(z)) + s_{n-1}(0) \\ &= d_{n+1}(s_n(z)). \end{align*} Because $s_n: C_n \to C_{n+1}$, we have $s_n(z) \in C_{n+1}$, and therefore $z \in \operatorname{im} d_{n+1} = B_n(C_\bullet)$. Hence \begin{align*} Z_n(C_\bullet) \subseteq B_n(C_\bullet). \end{align*}[/step]

[guided]The goal in degree $n$ is to prove that the homology group \begin{align*} H_n(C_\bullet) = Z_n(C_\bullet)/B_n(C_\bullet) \end{align*} is zero. This happens exactly when every cycle is already a boundary. Let $z \in Z_n(C_\bullet)$. By definition of $Z_n(C_\bullet)$, this means $z \in C_n$ and \begin{align*} d_n(z) = 0. \end{align*} The contractibility hypothesis gives homomorphisms $s_k: C_k \to C_{k+1}$ for all $k \in \mathbb{Z}$ satisfying \begin{align*} d_{k+1} \circ s_k + s_{k-1} \circ d_k = \operatorname{id}_{C_k}. \end{align*} Using this identity with $k = n$ and evaluating at $z \in C_n$, we obtain \begin{align*} z &= \operatorname{id}_{C_n}(z) \\ &= d_{n+1}(s_n(z)) + s_{n-1}(d_n(z)). \end{align*} Since $z$ is a cycle, $d_n(z) = 0$. Since $s_{n-1}: C_{n-1} \to C_n$ is a group homomorphism, it sends $0 \in C_{n-1}$ to $0 \in C_n$. Therefore \begin{align*} z &= d_{n+1}(s_n(z)) + s_{n-1}(0) \\ &= d_{n+1}(s_n(z)). \end{align*} Now $s_n(z) \in C_{n+1}$ because $s_n: C_n \to C_{n+1}$. Thus $z$ lies in the image of $d_{n+1}: C_{n+1} \to C_n$, which is precisely $B_n(C_\bullet)$. We have proved \begin{align*} Z_n(C_\bullet) \subseteq B_n(C_\bullet). \end{align*}[/guided]

[step:Conclude that the homology group vanishes in every degree] From the two inclusions \begin{align*} B_n(C_\bullet) \subseteq Z_n(C_\bullet) \quad \text{and} \quad Z_n(C_\bullet) \subseteq B_n(C_\bullet), \end{align*} we get \begin{align*} Z_n(C_\bullet) = B_n(C_\bullet). \end{align*} Therefore \begin{align*} H_n(C_\bullet) = Z_n(C_\bullet)/B_n(C_\bullet) = Z_n(C_\bullet)/Z_n(C_\bullet) = 0. \end{align*} Since $n \in \mathbb{Z}$ was arbitrary, $H_n(C_\bullet) = 0$ for every $n \in \mathbb{Z}$. [/step]