This post states that we can examine the deck transformations of the covering space $S^2 \rightarrow \mathbb{R}P^2 \rightarrow X$ to see that only the identity and antipodal maps are deck transformations. I am not sure how to show this is true. The homeomorphism is clear to me once I prove this.
If $q: S^2 \rightarrow \mathbb{R}P^2$ is the quotient covering map, and $p: \mathbb{R}P^2 \rightarrow X$ is any covering map, then a deck transformation $\phi: S^2 \rightarrow S^2$ for $p \circ q$ satisfies $$p \circ q = (p \circ q) \circ \phi.$$ For any $x \in S^2$, we have $p([x]) = p([\phi(x)])$ where $[x] = q(x)$ denotes an equivalence class in $\mathbb{R}P^2$. We don't have any further assumptions on $p$, so I don't know how to proceed from here. I can see that deck transformations with respect to $q$ are just the identity or antipodal maps, but composing with $p$ messes me up.
Here is a proof based on examining covering transformations of $p\circ q$. The group $G$ of covering transformations of $p\circ q$ contains the antipodal map. In order to show that $p$ is 1-1, it suffices to prove that $G$ has order 2. For each continuous map $f: S^2\to S^2$ define the induced map of homology group: $$ f_{*,i}: H_i(S^2; {\mathbb Q})\to H_i(S^2; {\mathbb Q}). $$
Let $G_0< G$ denote the index 2 subgroup consisting of covering transformations preserving orientation of $S^2$, i.e. inducting the identity map $f_{*,2}$.
For an element $f\in G_0$ compute its Lefschetz number $\Lambda_f$, $$ \Lambda_f= Tr(f_{*,0}) - Tr(f_{*,1}) + Tr(f_{*,2})= 1 -0+1=2. $$ Hence, by the Lefschetz theorem, $f$ has a fixed point in $S^2$. Since $f$ is a covering transformation, it follows that such $f$ is the identity map, i.e. $G_0=1$, i.e. $G$ has order $2$, i.e. $p$ is 1-1.