$\newcommand{\R}{\mathbf R}$ 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.