Use the transfer sequence for the covering \(X \times S^\infty \to X \times \mathbb{RP}^\infty\) to produce isomorphisms \(H_n(X \times \mathbb{RP}^\infty; \mathbb{Z}_2) \approx \bigoplus_{i \leq n} H_i(X; \mathbb{Z}_2)\) for all \(n\).
Solution
Put \[ E=X\times S^\infty, \qquad B=X\times\mathbb{RP}^\infty, \] and use \(\mathbb Z_2\) coefficients. The projection \(p:E\to B\) is a double cover, so its transfer sequence is \[ \cdots\to H_n(B)\overset{\tau_*}{\longrightarrow}H_n(E) \overset{p_*}{\longrightarrow}H_n(B) \overset{\delta}{\longrightarrow}H_{n-1}(B)\to\cdots . \] Since \(S^\infty\) is contractible, projection to the first factor gives \[ H_n(E)\cong H_n(X). \]
1. Under this identification, \(p_*:H_n(X)\to H_n(B)\) is split injective.
Choose \(e_0\in S^\infty\) and write \(b_0=[e_0]\in\mathbb{RP}^\infty\). After deforming \(E\) to \(X\times\{e_0\}\), the covering map becomes the inclusion \[ j:X\to X\times\mathbb{RP}^\infty, \qquad j(x)=(x,b_0). \] The projection \[ q:B\to X \] satisfies \(qj=\operatorname{id}_X\), so \(j_*=p_*\) is split injective.
2. The transfer map\[ \tau_*:H_n(B)\to H_n(E)\cong H_n(X) \]is zero for every \(n\).
Let \(r:E\to X\) be projection to the first factor. On singular chains, \(\tau\) sends a simplex in \(B\) to the sum of its two lifts. The two lifts have the same projection to \(X\), hence \[ r_\#\tau(\sigma)=2\,q_\#(\sigma)=0 \] over \(\mathbb Z_2\). Thus \(r_*\tau_*=0\). Since \(r_*:H_n(E)\to H_n(X)\) is an isomorphism, \(\tau_*=0\).
3. Hence for every \(n\) the transfer sequence breaks into a split short exact sequence\[ 0\longrightarrow H_n(X) \overset{p_*}{\longrightarrow}H_n(B) \overset{\delta}{\longrightarrow}H_{n-1}(B) \longrightarrow0. \]
By 2, exactness makes \(\delta\) surjective. By 1, \(p_*\) is injective, and exactness gives \(\ker\delta=\operatorname{im}p_*\). The splitting of \(p_*\) from 1 splits the short exact sequence.
4. Inductively,\[ \boxed{ H_n(X\times\mathbb{RP}^\infty;\mathbb Z_2) \cong \bigoplus_{i=0}^{n}H_i(X;\mathbb Z_2). } \]
From 3, \[ H_n(B)\cong H_n(X)\oplus H_{n-1}(B). \] For \(n=0\), \(H_0(B)\cong H_0(X)\) because \(\mathbb{RP}^\infty\) is connected. Iterating the displayed recurrence yields the stated direct sum.