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An immersion $M^4\to {\mathbb{R}}^5$ would inject $TM$ into the trivial rank-five bundle. Taking its orthogonal complement gives a normal real line bundle $\nu$ with $TM\oplus \nu \cong {\epsilon }^5$. The Whitney product formula for Stiefel–Whitney classes requiresThe real line bundle need not be orientable, but it has no Stiefel–Whitney classes above degree one.

Let $a,b$ be the pulled-back degree-one generators from the two real projective planes. Their mod-two cohomology ring, by the Künneth theorem, is ${\mathbb{F}}_2[a,b]/(a^3,b^3)$. To compute the tangent classes, at a line $L$ the projective tangent space is $\mathrm{Hom}(L,L^{\perp })$. Adding $\mathrm{Hom}(L,L)\cong \mathbb{R}$ gives $\mathrm{Hom}(L,{\mathbb{R}}^3)\cong 3L$, with the real line identified with its dual using a metric. Thus the stabilized tangent bundle of real projective space gives $T{\mathbb{RP}}^2\oplus {\epsilon }^1\cong 3\gamma$ andThe tangent bundle of the product is the direct sum of the pulled-back tangent bundles. Since $(1+a+a^2)(1+a)=1$ in this truncated ring, and similarly for $b$,The degree-two class $ab$ is nonzero by the product-ring basis. It cannot be a class of the rank-one bundle $\nu$. Equivalently $w_1(TM{)}^2+w_2(TM)=ab\ne 0$, violating the codimension-one Stiefel–Whitney immersion obstruction. Hence ${\mathbb{RP}}^2\times {\mathbb{RP}}^2$ admits no immersion in ${\mathbb{R}}^5$.