The statement is false for $n= 2$, as I'll show, with respect to both distances you are considering on $G(n,B(H))$ (this easily implies that it is also false for any $n>2$; on the other hand for $n=1$ the statement is true according to your definition of Fredholm subspace, as it reduces to the fact that Fredholm operators are an open subset of $B(H)$. )

Let $R$ be the surjective shift operator (the adjoint of the injective shift operator). Recall that for any $\lambda\in \mathbb{C}$ the operator $ R-\lambda$ is surjective and Fredholm of index $1$ if $|\lambda|<1$, it is not Fredholm if $|\lambda|=1$, and it is invertible if $|\lambda|>1$. It will be of use a countable direct sum  of operators $\lambda_j R$; to do it more nicely, we  can realize it on a suitable Hilbert decomposition of $\ell^2$ into countably many infinite dimensional subspaces.

So let $\alpha:=(\alpha_k)_{k\ge1}$ be a bounded sequence of complex numbers.
Consider the bounded operator $L^\alpha$ on $H:=\ell^2(\mathbb{Z}_+)$ defined on the standard orthonormal bases $(e_k)_{k\ge1}$ by
$$L^\alpha e_k = \alpha_k e_{2k}\ .$$
Assume further that  the sequence $\alpha$ satisfies $\alpha_{2k}=\alpha_k$ for all $k\ge 1$. Then $H$  splits in a Hilbert sum of a countable family of $L^\alpha$-invariant subspaces, $H=\bigoplus_{j\in\mathbb{N}} H_j $, with $H_j:=\operatorname {Span}(e_{(2j+1)2^n}: n\ge0)$. Moreover, for any $j\ge0$ the operator $L^\alpha_{\ |H_j}$ on $H_j$ is unitary equivalent to $\alpha_{2j+1}R$ on $\ell^2$. Thus for any $\lambda\in\mathbb{C}$ the operator $L^\alpha-\lambda I$ is 
unitary equivalent to the direct sum $\bigoplus_{j\in\mathbb{N}} \big(\alpha_{2j+1}R-\lambda\big)$.  The $j$-th component $\alpha_{2j+1}R-\lambda$ is 
either non-Fredholm, or surjective Fredholm of index $1$, or invertible according whether $|\lambda|=|\alpha_{2j+1}|$, $|\lambda|<|\alpha_{2j+1}|$, respectively $|\lambda|>|\alpha_{2j+1}|$, in which case $\big\|\big(\alpha_{2j+1}R-\lambda\big)^{-1}\big\|\le\frac{1}{|\lambda|-|\alpha_{2j+1}|} .$


Therefore  the operator $L^\alpha-\lambda I$ is Fredholm of index
$$\mathrm{ind}(L^\alpha-\lambda I)=\mathrm{card}  \{j\ge0\ : |\alpha_{2j+1}| > |\lambda|  \}\ , $$  


provided $|\alpha_{2j+1}|$ is bounded away from $|\lambda|  $  and $|\alpha_{2j+1}|<|\lambda|  $ for all but finitely many $j$, say $m$:
indeed, in this case $L^\alpha-\lambda I$ is the direct sum of a bounded family of invertible operators whose inverses have bounded norms, plus   $m$ Fredholm operators of index $1$, thus the direct sum of an invertible operator and a Fredholm operator of index $m$. 

 
As a consequence, the $2$-dimensional subspace $V^\alpha$ generated by the operator $L^\alpha$ and the identity $I$ is a Fredholm subspace according to your definition, if  e.g. the sequence $|\alpha_{2j+1}|$  is eventually constant, and it is not, if e.g. it is strictly decreasing. Indeed, in the  former case  the index of Fredholm operators in $V^\alpha$ can only assume finitely many values, while in the latter case  $V^\alpha$ possesses Fredholm operators of any index. On the other hand, for bounded sequences $\alpha$ and $\beta$ one has $\|L^\alpha-L^\beta\|\le |\alpha-\beta|_\infty$ for the operator norm, so that $V^\alpha$ depends continuously from $\alpha $ (and a fortiori, if you adopt the inner product norm deduced from $\langle\ ,\ \rangle$, which is weaker). Since an eventually constant sequence is arbitrarily close to strictly decreasing sequences in the $\ell_\infty$ norm, we conclude that Fredholm subspaces of $G(2,B(H))$ are not an open subset.