Does the Poincare homology sphere smoothly embed in $\mathbb{C}P^2$?
If so I guess one way to see this would be to have a handle decomposition for $\mathbb{C}P^2$ such that a subset of the 2 handles, together with all of the 1 handles is a surgery description of the Poincare homology sphere (after changing all of the dots on the 1-handles to zeros). I can't seem such a description of $\mathbb{C}P^2$.
I do not know how one would go about proving that such an embedding does not exist.
No. Splitting $\Bbb{CP}^2$ into two pieces $X_1, X_2$ along the Poincare homology sphere is such that $H_*(X_1) \oplus H_*(X_2) = H_*(\Bbb{CP}^2)$ for $* < 3$ (run Mayer-Vietoris). In particular we split the intersection form. Therefore one of the two pieces is a homology ball and the other is positive-definite. $\Sigma(2,3,5)$ can not bound the first piece, as it never bounds any negative definite manifold with diagonal intersection form by Donaldson's theorem: if it did, then capping off the $-E8$-manifold (a smooth manifold with intersection form $-E8$ and boundary $-\Sigma(2,3,5)$) gives a smooth closed oriented 4-manifold with negative definite, non-diagonalizable intersection form.
In a related but different direction, Proposition 3.4 of this great paper shows that the minimal $k$ such that $\Sigma(2,3,5)$ embeds in $\#^k S^2 \times S^2$ is $k = 8$.