I am trying to understand this problem. All answers I find say more or less the same thing, the canonical projection is an open map, so we are mapping a compact (and hence closed) thing into an open set and hence we have a clopen set, but that must be $R^n$ contradiction since $R^n$ is not compact.
But all I get from the definition of an open map is, if the domain is open the image is open. I don't see why the canonical projection map MUST map something into an open set.
Like, I can project the closed ball in $R^3$ down into the closed disk in $R^2$. This is a mapping between 2 closed sets but it's also just removing the last coordinate from elements in the domain, which if I understand correctly is literally what the canonical projection is.
I see no reason or argument that explains why having an open map suddenly guarantees that the image is open.
i.e. if the formal statement is $f$ open map $\iff$ $D$ is open $\implies$ $Im(f)$ open.
That says nothing about the behaviour of $f$ on closed sets. Like I can maybe find a closed set that is a subset of the domain of $f$, so the function must be defined in closed sets as well.
In short, I don't see why this argument is true:
Submersions are open maps; but the image of $M$ is compact in a Hausdorff space, and hence closed as well. So it's a clopen nonempty set. Since Rn is connected, it's the whole thing. But then $R^n$ is the quotient of a compact space, so it's compact, which is not true.