$\def\RR{\mathbb{R}}\def\CC{\mathbb{C}}\def\PP{\mathbb{P}}$I think the answer is no. Topology isn't my strength, so check this argument.
Let $\infty$ be the point $[0:0:1]$ in $\CC \PP^2$ and put $U = \CC\PP^2 \setminus \{ \infty \}$. Note that $\{ [z_1:z_2:0] \} \cong \CC\PP^1 \cong S^2$, I'll always identify $S^2$ with this particular subset of $U$.
At a point $z=[z_1:z_2:0]$ in $S^2$, let $L(z)$ be the line of $\CC\PP^2$ through $z$ and $\infty$. We have the decomposition of tangent spaces $$T_z U = T_z S^2 \oplus T_z L(z).$$ This gives an equality of vector bundles: $$TU|_{S^2} \cong TS^2 \oplus \mathcal{L}$$
Now suppose that $U$ embeds into $\RR^5$. Then we can choose a normal direction to $U$, and get $$T\RR^5|_U \cong TU \oplus \underline{\RR}$$ where $\underline{\RR}$ denotes the trivial bundle. And $T\RR^5$ is trivial, so we can rewrite this as $$\underline{\RR}^5|_U \cong TU \oplus \underline{\RR}.$$ Restricting to $S^2$, $$\underline{\RR}^5|_{S^2} \cong TS^2 \oplus \mathcal{L} \oplus \underline{\RR}.$$ But $TS^2$ is stably trivial and $\mathcal{L}$ represents the nontrivial class in $K_0(S^2)$, a contradiction.