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. We have $$\pi_3(\Bbb RP^3\times \Bbb RP^3) = \pi_3(\Bbb RP^3)^2 = \pi_3(S^3)^2= \Bbb Z^2$$ but $\pi_3(\Bbb RP^6) = 0$, so they cannot be homeomorphic.
