I want to prove that $LS(\mathbb{RP}^2)=3$, where $$LS(X) = \min{\lbrace k \in \mathbb{N} \hspace{3pt} | \hspace{3pt} \text{there exists an open cover }U_{1},...,U_{k} \text{ s.t. } i_{l}:U_{l} \hookrightarrow X \text{ is null-homotopic} \rbrace}$$
It was easy enough to prove that $LS(\mathbb{RP^2})\leq 3$, by finding three orthogonal great circles on the sphere $S^2$, taking small belts around them $B_{1}$, $B_{2}$ and $B_{3}$, and setting $U_{i} = p(S^2 \setminus B_{i})$, where $p:S^2 \to \mathbb{RP}^2$ is the standard covering map.
However, I can't find a proof for $LS(\mathbb{RP^2}) \geq 3$.
I start by supposing that there exist open $A$ and $B$ whose inclusions are null-homotopic. Then $(\mathbb{RP}^2; A, B)$ is an excisive triad (https://en.wikipedia.org/wiki/Excisive_triad), so by excision we have that $j:(A, A \cap B) \hookrightarrow (\mathbb{RP^2}, B)$ induces an isomorphism in the homology, so $H_{n}(A, A \cap B) \cong H_{n}(\mathbb{RP^2}, B)$. Also, from the long exact sequence of $(\mathbb{RP}^2, B)$ and null-homotopic-ness of $i_{l}$ we have that the following sequences are short and exact (for all $n \in \mathbb{N}$):
$$0 \to H_{n}(\mathbb{RP^2}) \to H_{n}(\mathbb{RP^2}, B) \to H_{n-1}(B) \to 0.$$
However, I can't connect these two facts, because I have no idea how to find $H_{n}(A, A \cap B)$, $H_{n}(A)$, $H_{n}(B)$ or $H_{n}(B, A \cap B)$.
Is it possible to derive a contradiction from here?