I consider the set $ T:=GL_n(\mathbb{R})\subseteq \mathbb{R}^{n,n}=:X $ with the metirc $$d:\mathbb{R}^{n,n}\times \mathbb{R}^{n,n},(A,B)\mapsto \text{rank}(A−B).$$
I want to show that $ T $ is not open in $ (X,d) $. I use the fact: $ T\subseteq X $ open in $ (X,d) $ iff $$ \ \forall A\in \ T\ \exists \varepsilon>0:B_{\varepsilon}(A)\subseteq T. $$
So I want to show: $$ \exists A\in \ T\ \forall \varepsilon>0:B_{\varepsilon}(A)\nsubseteq T. $$
At first I take $ 1<\varepsilon $ and chose $ B\in \mathbb{R}^{n,n} $ with the first $ n-1 $ colums of $ A\in T $ and the n-th colum is just the zero vector. So indeed we get $ \text{rank}(A-B)=1<\varepsilon $ such that $ B\in B_{\varepsilon}(A) $ and $ B $ is not invertible.
But in my opinion the values $ 0<\varepsilon\leq 1 $ are gonna bring me in trouble. If I consider those values $ \varepsilon $ and a matrix $ M\in B_{\varepsilon}(A) $ then I get $ \text{rank}(A-M)<\varepsilon\leq 1 $. But then it applies $ A=M $ which means $ M $ invertible. So for this special case of $ \varepsilon $ I cannot find a non-invertible matrix.