You can also prove this using homology, but it's somewhat more effort. The basic idea is that, if $B$ is the boundary circle and $r:M\rightarrow M$ is a retraction onto $B$ then the inclusion map also induces an injection $i_*:H_1(B)\rightarrow H_1(X)$ (to see this, apply $r_*$ to $i_*\alpha=i_*\beta$). Thus we have an exact sequence
$$
0\rightarrow H_1(B)\xrightarrow{i_*}H_1(M)\xrightarrow{q_*} H_1(M,B)\rightarrow 0
$$
coming from the reduced long exact sequence for the pair $(M,B)$. We know that $H_1(B)$ and $H_1(M)$ are both $\mathbb Z$ (because $B=S^1$ and $M$ deformation retracts onto its central circle) and, since $(M,B)$ is a good pair, $H_1(M,B)\cong H_1(M/B)$. But $M/B=\mathbb R\mathbb P^2$, as can be seen by their cell decompositions, where the pink indicates the boundary circle $B$:

thus $H_1(X,B)\cong \mathbb Z/2\mathbb Z$. The fact that $r$ is a retraction means that the above sequence splits, as $r_*:H_2(M)\rightarrow H_2(B)$ composed with $i_*$ is the identity. This is a contradiction.