= Solution
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 $TM\oplus\nu\cong\varepsilon^5$. The <Whitney product formula for Stiefel–Whitney classes> requires
$$
w(TM)w(\nu)=1.
$$
The <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 $\operatorname{Hom}(L,L^\perp)$. Adding $\operatorname{Hom}(L,L)\cong\mathbb R$ gives $\operatorname{Hom}(L,\mathbb R^3)\cong3L$, with the real line identified with its dual using a metric. Thus the <stabilized tangent bundle of real projective space> gives $T\mathbb{RP}^2\oplus\varepsilon^1\cong3\gamma$ and
$$
w(T\mathbb{RP}^2)=(1+a)^3=1+a+a^2.
$$
The <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 $b$,
$$
w(TM)^{-1}=(1+a)(1+b)=1+a+b+ab.
$$
The degree-two class $ab$ is nonzero by the product-ring basis. It cannot be a class of the rank-one bundle $\nu$. Equivalently $w_1(TM)^2+w_2(TM)=ab\ne0$, violating the <codimension-one Stiefel–Whitney immersion obstruction>. Hence \b[$\mathbb{RP}^2\times\mathbb{RP}^2$ admits no <immersion> in $\mathbb R^5$].
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