The statement is false for $n= 2$, as I'll show, with respect to both distances you consider 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 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 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 $L^\alpha$ on $H_j$ is isomorphic to $\alpha_{2j+1}R$ on $\ell^2$. Thus for any $\lambda\in\mathbb{C}$ the operator $L^\alpha-\lambda I$ splits into components on $H_j$ that are either invertible, or surjective Fredholm of index $1$, or non-Fredholm, according whether $|\lambda|>|\alpha_{2j+1}|$, $|\lambda|<|\alpha_{2j+1}|$, respectively $|\lambda|=|\alpha_{2j+1}|$. Therefore,  if $|\lambda|\neq|\alpha_{2j+1}|$ for all $j\ge0$, the operator $L^\alpha-\lambda I$ is semi-Fredholm of index
$$\mathrm{ind}(L^\alpha-\lambda I)=\mathrm{card}  \{j\ge0\ : |\alpha_{2j+1}| > |\lambda|  \}\ . $$

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. 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 prod norm deduced from $\langle\ ,\ \rangle$, which is weaker). Since eventually constant sequences may be very close to strictly decreasing sequences in the $\ell_\infty$ norm, we conclude that Fredholm subspaces of $G(2,B(H))$ are not open subset.