I am trying to determine whether or not $M := \mathbb{R}P^4/\mathbb{R}P^2$ is a manifold. I believe it is clear that $M$ is second-countable. However, I'm not not sure about Hausdorffness and being locally homeomorphic to $\mathbb{R}^d$ for some $d$.
My idea to show that $M$ is indeed a manifold was to find some "nice" (i.e. a manifold) covering space $\widetilde{M}$ of $M$ and then argue as usual that $M$ itself is a manifold; but I have not had any luck here. Perhaps there is some known space homeomorphic to $\mathbb{R}P^n/\mathbb{R}P^m$ for general $n \geq m$? On the other hand, I'm not sure where to start if I were to disprove the claim.
I feel like I'm missing something obvious. I am grateful for any hints.
Moishe Kohan's comment is a great way to directly show that this space $X = \mathbb{R}P^4/\mathbb{R}P^2$ is not a manifold, but here's a way to show that it actually can't even have the homotopy type of a (topological) manifold.
Consider the standard CW structure on $\mathbb{R}P^4$ with one cell in each dimension $\leq 4$ so that the $k$-skeleton of $\mathbb{R}P^4$ is $\mathbb{R}P^k$. Thus to form $X$, one collapses the $2$-skeleton. The resulting space has one $0$-cell, one $3$-cell, and one $4$-cell (attached to the $3$-skeleton, which is necessarily $S^3$, by a map of degree 2).
Now consider the cellular chain complex for $X$ with this CW structure, as well as the cellular cochain complex for $X$, with $\mathbb{Z}_2$ coefficients (we could use integer coefficients but it makes the answer a little less pretty below). Computing the (co)homology of these chain complexes results in
$$H_j(X;\mathbb{Z}_2) \cong H^j(X; \mathbb{Z}_2) \cong \begin{cases} \mathbb{Z}_2, &j = 0, 3, 4 \\ 0, &\text{otherwise.} \end{cases}$$
This contradicts Poincare Duality with $\mathbb{Z}_2$ coefficients, which must hold if $X$ is to have the homotopy type of a manifold.