Homology of quotient of 3-sphere by identifying antipodal points on equator

Your $H_3$ calculation is correct because $H_2(A\cap B)=0$. The $H_2$ calculation is a little fishy. Here is the relevant fragment of the MV sequence: $$0\to H_2(X)\to H_1(A\cap B) \to H_1(A)\oplus H_1(B),$$ which turns into $$0\to H_2(X) \to \mathbb Z_2\to \mathbb Z_2\oplus\mathbb Z_2.$$ There is no reason to conclude that $H_2(X)=\mathbb Z_2$. Indeed, based on the topology, the map $\mathbb Z_2\to \mathbb Z_2\oplus\mathbb Z_2$ should be given by $x\mapsto (x,x)$, which is injective. Hence $H_2(X)=0$.

But now you have enough to figure out $H_1(X)$. This is easiest if you use reduced homology. Then, similarly to what you've written you get $$0\to \mathbb Z_2\to \mathbb Z_2\oplus\mathbb Z_2\to H_1(X)\to 0.$$ Thus $H_1(X)\cong (\mathbb Z_2\oplus\mathbb Z_2)/\langle (x,x)\rangle\cong \mathbb Z_2$.

An alternative approach is to use cellular homology. This is basically $\mathbb{RP}^2$ with two $3$-cells attached. The cellular chain complex is $$0\to \mathbb Z^2\overset{0}{\to} \mathbb Z\overset{\times 2}{\to}\mathbb Z\overset{0}{\to}\mathbb Z\to 0.$$ You can deduce these maps since you know what they are for $\mathbb{RP}^3$, and this is essentially $\mathbb{RP}^3$ with an additional $3$ cell attached in the same way. Taking the homology gives the same answer as Mayer-Vietoris.