TITLE: Looking For a Neat Proof of the Fact that the Grassmannian Manifold is Hausdorff
QUESTION [16 upvotes]: $\newcommand{\R}{\mathbf R}$
Let $V$ be an $n$-dimensional vector space and $k$ be an integer less than $n$. A $k$-frame in $V$ is an injective linear map $T:\R^k\to V$. Let the set of all the $k$-frames in $V$ be denoted by $F_k(V)$. It is clear that $F_k(V)$ is an open subset of $L(\R^k, V)$. 
Define a relation $\sim$ on $F_k(V)$ as follows: We write $S\sim T$ for two members $S$ and $T$ in $F_k(V)$ if and only if $\text{span }T=\text{span }S$.
It can be easily seen that $S\sim T$ if and only if there is a $\tau\in GL_k(\R)$ such that $T=S\circ \tau$.
The Grassmannian manifold $GR_k(V)$ is defined as the quotient space $F_k(V)/\sim$.
I know that the projection map $\pi:F_k(V)\to GR_k(\R)$ is an open map.

I am trying to prove that $GR_k(V)$ is a Hausdorff space.

I proved the above by noting that the above statement is just this: Given linearly independent lists $(u_1, \ldots, u_k)$ and $(v_1, \ldots, v_k)$ in $V$ which do not span the same subspace, there are neighborhoods $U_i$'s of $u_i$'s and $V_j$'s of $v_j$'s such that whenever $(u_1',\ldots, u_k')\in U_1\times \cdots\times U_k$ and $(v_1',\ldots, v_k')\in V_1\times \cdots\times V_k$, the lists $(u_1', \ldots, u_k')$ and $(v_1', \ldots, v_k')$ are linearly independent and do not span the same subspace.
I had a rather long proof of this. Basically I established that given hyperplanes $H$ and $K$ in $V$, there is a hyperplane $P$ in $V$ such that $P$ is "between" $H$ and $K$.
I am looking for a more direct approach than recasting the problem in the aformentioned way.
edit:
Also, I am trying to avoid the use of matrices and coordinates as much as possible.
Thanks.
EDIT.
I finally was able to put down the kind of proof I was looking for. Here it is.

REPLY [3 votes]: Another 'neat' proof that has not been mentioned already (I hope) goes as follows:
You can view $Gr(k,V)$ as equivalence classes of $n\times k$ matrices of full rank. Trivialised sets $U_\alpha$ are those matrices $X$ which give invertible $k\times k$ matrices $X_\alpha$ by selecting the rows $\alpha$. Here $\alpha$ is a multi-index of $k$ integers between $1$ and $n$. You would map these matrices to $\mathbb{R}^{k(n-k)}$ by changing coordinates to make the $\alpha$ submatrix the identity. It is 'easy' to see this is a cover with polynomial transition functions.
To show this space is Hausdorff, consider two points $A$, $B$ in $U_\alpha$, $U_\beta$ respectively, neither one being in the intersection. Define functions $h_{\alpha\beta}:U_\alpha \rightarrow\mathbb{R}$ and $h_{\beta\alpha}:U_\beta \rightarrow\mathbb{R}$ by $h_{\alpha\beta}(X):=det(X_{\beta}X_\alpha^{-1})$ and $h_{\beta\alpha}(X):=det(X_{\alpha}X_\beta^{-1})$ respectively. These are well defined and multiply to give $1$ on $U_\alpha\cap U_\beta$.
Since neither $A$ nor $B$ is in the intersection, they respectively map to $0$ by the functions. The claim is that $h_{\alpha\beta}^{-1}(-1,1)$ and $h_{\beta\alpha}^{-1}(-1,1)$ are disjoint open neighbourhoods for $A$ and $B$. To see they are disjoint, use the fact that $h_{\alpha\beta}$ and $h_{\beta\alpha}$ multiply to give $1$ on $U_\alpha \cap U _\beta$.