Let us first show the following result,
Theorem 1.1
\(\tilde{H}_n(SX) \cong \tilde{H}_{n - 1}(X)\)
Proof. Observe,
Lemma 1.2
\(H_n(X, A) \cong \tilde{H}_n(X \cup CA)\)
This follows from Excision since, \[ \tilde{H}_n(X \cup CA) \cong H_n(X \cup CA, CA) \cong H_n(X \cup CA \setminus \{p\}, CA \setminus\{p\}) \cong H_n(X, A) \] where the first isomorphism follows from \(CA\) being contractible!
Observe that since \(SX = CX \bigcup_{X} CX\), thus by the lemma we know that \(\tilde{H}_n(SX) \cong H_n(CX, X)\) and by the long reduced homology chain, \[ \ldots \longrightarrow \tilde{H}_n(CX) \longrightarrow H_n(CX, X) \longrightarrow \tilde{H}_{n - 1}(X) \longrightarrow \tilde{H}_{n - 1}(CX) \longrightarrow \ldots \] and since \(\tilde{H}_n(CX)\) is zero (since \(CX\) is contractible) we conclude the desired result!
Now let us explore the problem more generally, let \(C^kX\) be \(k\) copies of \(CX\) glued over \(X\).
Theorem 1.3
\(\tilde{H}_n(C^kX) \cong \bigoplus_{i = 1}^{k - 1} \tilde{H}_{n - 1}(X)\)
Proof. Let us decompose \(C^kX = CX \cup C^{k - 1}X\), then \(CX \cap C^{k - 1}X = X\), thus by the Mayer–Vietoris sequence we obtain, \[ \ldots \longrightarrow \tilde{H}_n(X) \longrightarrow \tilde{H}_n(C^{k - 1}X) \oplus \tilde{H}_n(CX) \longrightarrow \tilde{H}_n(C^kX) \longrightarrow \tilde{H}_{n - 1}(X) \longrightarrow \ldots \] since \(CX\) is contractible the \(\tilde{H}_n(CX)\) term disappears. Let us closely look at the map \(\tilde{H}_n(X) \longrightarrow \tilde{H}_n(C^{k - 1}X)\), clearly this induces the zero map (via the inclusion \(X \hookrightarrow C^{k - 1}X\)) since every cycle in \(X\) becomes contractible in \(C^{k - 1}X\), thus we obtain the short exact sequence, \[ 0 \longrightarrow \tilde{H}_n(C^{k - 1}X) \longrightarrow \tilde{H}_n(C^kX) \longrightarrow \tilde{H}_{n - 1}(X) \longrightarrow 0 \] now to apply the Split Lemma we try to construct the inverse map which we can do since for the first map simply consider the map that maps the last cone into the first one. Thus, \[ \tilde{H}_n(C^kX) \cong \tilde{H}_n(C^{k - 1}X) \oplus \tilde{H}_{n - 1}(X) \] thus by induction we obtain the desired result!
Remark 1.4
A much simpler approach can be obtained by observing that \(C^k X \simeq C^kX / C_1X\) (choosing an arbitrary cone \(C_1X\)) thus when collapsed will collapse \(X\) and thus \(C^k X \simeq \bigvee_{i = 1}^{k - 1} SX\), thus by the previous lemma we obtain the desired result.