You are not assuming, a priori that the isometry is differentiable, so appeals to Riemannian geometry wouldn't do you any good anyway. Likewise, appeals to double covers and the idea that one could 'lift' the distance function on $\mathbb{RP}^2$ to $S^2$ somehow are likely to get you into problems if you want to do it by purely elementary means.
I think that a reasonable approach is the following one: If you think of $\mathbb{RP}^2$ as the space of lines through the origin in $\mathbb{R}^3$ and denote by $[u]\in\mathbb{RP}^2$ the line spanned by a nonzero vector $u\in\mathbb{R}^3$, then the function $f:\mathbb{RP}^2\times \mathbb{RP}^2\to \mathbb{R}$ defined by
$$
f\bigl([u],[v]\bigr) = \frac{(u\cdot v)^2}{(u\cdot u)(v\cdot v)}\tag1
$$
is well-defined and is equivalent to specifying your distance function (since $\sin^2\bigl(\cos^{-1}r\bigr)= 1-r^2$). Thus, you are asking to show that any map $\phi:\mathbb{RP}^2\to \mathbb{RP}^2$ that satisfies $$f\bigl(\phi([u]),\phi([v])\bigr) = f\bigl([u],[v]\bigr)\tag2$$ for all $u,v\in\mathbb{RP}^2$ is of the form $\phi([u]) = [Au]$ for some $A\in\mathrm{SO}(3)$. (Since $[-u]=[u]$, you don't need $\mathrm{O}(3)$, as it does not act effectively on $ \mathbb{RP}^2$.)
Now, since $\mathrm{SO}(3)$ already acts transitively on triples of orthogonal lines and preserves $f$, it follows that it's enough to understand the isometries that fix three given points $[u_1],[u_2],[u_3]$ that satisfy $f\bigl([u_i],[u_j]\bigr)= \delta_{ij}$, and you might as well assume that these points are $[u_1] = [1,0,0]$, $[u_2]=[0,1,0]$, and $[u_3]=[0,0,1]$. Since there are 4 points that are at equal distance from all three of these points, namely $[x_1,x_2,x_3]$,
where $x_i^2 = 1/3$, we can, by applying a diagonal element of $\mathrm{SO}(3)$, further assume that $\phi$ fixes the point $[u_4] = [x,x,x]$ where $x^2 = 1/3$.
Now, the task is to show that this forces such a $\phi$ to fix all points.
Take any point $[v] = [x,y,z]\in\mathbb{RP}^2$ and assume, as we can, that $x^2+y^2+z^2 = 1$. Let $[w] = \phi([v]) = [p,q,r]$, where $p^2+q^2+r^2=1$. Since $\phi$ fixes $[u_i]$ and satisfies (2), we know that
$$
p^2 = x^2,\quad q^2 = y^2,\quad r^2 = z^2,\quad (p+q+r)^2=(x+y+z)^2.\tag3
$$
If, say $x=0$, then $p=0$, and the remaining equations $q^2=y^2$, $r^2=z^2$
and $(q+r)^2=(y+z)^2$ easily imply that $(q+ir)^2 = (y+iz)^2$, so $[0,q,r]=[0,y,z]$, so $\phi([v])=[v]$.
A similar argument shows that, if either $y=0$ or $z=0$, then $\phi([v]) = [v]$.
Since points of the form $[\cos\theta,\sin\theta,0]$ are now known to be fixed by $\phi$, it follows that $(\cos\theta\,x+\sin\theta\,y)^2= (\cos\theta\,p+\sin\theta\,q)^2$ for all $\theta$, and hence, $xy=pq$. Similarly, $xz=pr$ and $yz=qr$. This, coupled with the above equations, implies that $(p,q,r) = \pm(x,y,z)$, i.e., that $\phi([v])=[v]$, as was to be shown.
N.B.: Note that one doesn't need to assume that $\phi$ is invertible (or even surjective) in order for this proof to work. (While these are not hard to show directly, they would be extra steps that would require some care and a higher level of sophistication, since they seem to require notions of continuity and compactness that are, themselves, above the merely algebraic.)
