To show that the Stiefel Manifold is compact and Hausdorff I have been given the following space:
$$X=V_m(\mathbb{R}^n)=\{(v_1,\cdots,v_m)\in \mathbb{R}^n \times\cdots\times\mathbb{R}^n \ \vert \ v_1,\cdots,v_m \ \text{are linearly independent}\}$$
And we have defined the following equivalence relation $\sim$
$$(v_1,\cdots,v_m)\sim(v'_1,\cdots,v'_m) \ \text{iff}\ (v_1,\cdots,v_m) \ \text{and} \ (v'_1,\cdots,v'_m) \ \text{span the same}\ m-\text{plane}$$
I have to show that $X/\sim \ $ is compact, Hausdorff and each of it's points has a neighbourhood homeomorphic to $\mathbb{R}^{m(n-m)}$
Now, X has the induced topology from the standard topology on $\mathbb{R}^{mn}$. But I have no idea on how to proceed. I have also been provided the following hint, but I don't know how to use it.
Hint: if $\mathbb{R}^n=V \bigoplus W$ then one can view $V$ as the graph of a linear mapping $V \rightarrow W$, namely, the zero mapping. What other subspaces of $\mathbb{R}^n$ of dimension $\dim V$ can be regarded as graphs of linear mappings $V \rightarrow W$
Any kind of assistance is helpful! Especially if someone can explain to me how the hint might relate to the question. 
 A: Here's my solution:
We have 
$$X=V_m(\mathbb{R}^n)=\{(v_1,\cdots,v_m)\in \mathbb{R}^n \times\cdots\times\mathbb{R}^n \ \vert \ v_1,\cdots,v_m \ \text{are linearly independent}\}$$ as a topological subspace with the induced topology from $\mathbb{R}^{m \times n}$
We introduce the function 
$$\text{gs}: \mathbb{R}^n \times\cdots\times\mathbb{R}^n \rightarrow \mathbb{R}^n \times\cdots\times\mathbb{R}^n \\ (v_1,\cdots,v_m) \mapsto \text{gs}(v_1,\cdots,v_m)$$ where $\text{gs}(v_1,\cdots,v_m)$ is the Gram-Schmidt Orthonormalisation of $(v_1,\cdots,v_m)$
We now define the subspace $X_* \subset X$:
$$X_*=\{\text{gs}(v_1,\cdots,v_m) \ \vert \ (v_1,\cdots,v_m) \in V_m(\mathbb{R}^n)\}$$
which is our familiar compact Stiefel Manifold.
We have defined the following equivalence relation $\sim$
$$(v_1,\cdots,v_m)\sim(v'_1,\cdots,v'_m) \ \text{iff}\ \ \text{Span}(v_1,\cdots,v_m) = \text{Span}(v'_1,\cdots,v'_m)$$
and we have the map $\pi:X \rightarrow X/\sim$
We clearly see that $\pi(X_*)=X/\sim \implies \pi|_{X_*}:X_* \rightarrow X/\sim$ is a continuous surjective function and hence $X/\sim$ is compact.
We know that the $X/\sim$ is a manifold of dimension $m(n-m)$ hence any open  neighbourhood is homeomorphic to $\mathbb{R}^{m(n-m)}$
