View $\mathbb RP^2$ as $\mathbb RP^1$ attached with a 2-dimensional open unit ball $e^2$. Let $U$ be an open neighborhood of $\mathbb RP^1$ and $V$ be $e^2$. Then $U \cap V$ deformation retracts onto a circle $S^1$. By Mayer-Vietoris, we have the following seuqence:
$H^{0}(\mathbb RP^1) \oplus H^{0}(e^2) \xrightarrow{\alpha} H^{0}(S^1) \xrightarrow{\beta} H^1(\mathbb RP^2) \to H^1(\mathbb RP^1) \oplus H^1(e^2) \xrightarrow{\gamma} H^1(S^1) \xrightarrow{\lambda} H^2(\mathbb RP^2) \to H^2(\mathbb RP^1)\oplus H^2(e^2) $
I am not sure about how to calculate $\alpha, \beta, \gamma, \lambda$:
$\alpha$:Both $H^{0}(\mathbb RP^1)$ and $H^{0}(e^2)$ are generated by the constant function with value 1 on them, so a basis would be $(1,0)$ and $(0,1)$ which are mapped to the constant function with value $-1$ and $1$. So $\alpha$ sends $(m,n)$ to $n-m$ which is surjective.
$\gamma$: $H^1(\mathbb RP^1)$ is generated by a volume form $\omega$ of $RP^1$, $(\omega, 0)$ is mapped to $(-\omega,0)$, then the map is given by $(m,0) \to -m$, being surjective
But this cannot be true, since if $\gamma$ is surjective, then the kernel of $\lambda$ would be $\mathbb R$, then $H^2(\mathbb RP^2) \cong 0$ which is cleraly wrong.