Is $\mathbb{R}P^6 = \mathbb{R}P^3 \times \mathbb{R}P^3$? Question is in the title. I know that $S^n$ is not $S^1 \times \cdots \times S^1$, but it is true that $\mathbb{R}^n = \mathbb{R}^1 \times \cdots \times \mathbb{R}^1$.

No. The universal covers of the two spaces are, respectively, $\Bbb S^6$ and $\Bbb S^3 \times \Bbb S^3$, but these spaces are not homeomorphic, as, e.g., their $3$rd homology groups differ: $H_3(\Bbb S^3 \times \Bbb S^3) = \Bbb Z \oplus \Bbb Z$ but $H_3(\Bbb S^6) = 0$.
Another way to see that the proposed "product rule" for $\Bbb R \Bbb P^k$ fails is to observe that $\Bbb{RP}^1 \times \Bbb{RP}^1 \cong \Bbb S^1 \times \Bbb S^1$ is orientable but $\Bbb{RP}^2$ is not.
