Let $E$ be a Banach space. Recall that the collection of all closed linear subspaces of $E$ can be turned into a metric space in a number of ways. In particular, consider the notion of a **gap**: if $G$ and $H$ are subspace of $E$, then
$$g(G,H)=\max\{\sup\limits\_{g\in \partial B\_{G}} d(g, H),~\sup\limits\_{h\in \partial B\_{H}} d(h, G)\},$$
where by $\partial B\_{G}$ and $\partial B\_{H}$ we mean the intersection of the unit sphere with $G$ and $H$, respectively. Note, that $g(G,H)\le h(G,H)\le 2g(G,H)$, where $h(G,H)$ is the Hausdorff distance between $\partial B\_{G}$ and $\partial B\_{H}$.



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> Let $F\subset G$ be subspaces of $E$ and let $\varepsilon>0$. Does there exist $\delta>0$ such that every subspace $H$ with $g(G,H)<\delta$ contains a sub-subspace $J\subset H$ with $g(F,J)<\varepsilon$?
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If a subspace $F$ is contained in a subspace $G$, and $H$ is close to $G$, can we choose a subspace of $H$ close to $F$?
The answer is "No". You can derive this from Lemma 5.9 and Proposition 5.3 in my paper Ostrovskiĭ, M. I. [Topologies on the set of all subspaces of a Banach space and related questions of Banach space geometry](https://doi.org/10.1080/16073606.1994.9631766). Quaestiones Math. 17 (1994), no. 3, 259–319. In that Lemma a collection of subspaces $G\_\varepsilon$, $\varepsilon\in(0,1)$ is constructed in the space $X\oplus X/Y$ which converges to $Y\oplus X/Y$ with respect to gap, and is such that all $G\_\epsilon$ are isomorphic to $X$.


We get a counterexample in cases where $X/Y$ does not admit an isomorphic embedding into $X$ (such examples are well-known, e.g. $X=\ell\_1$ and $X/Y=c\_0$). In fact, if there would be a subspace $U$ in $G\_\varepsilon$, which is close to $X/Y$, since $X/Y$ is complemented in the whole space, by Berkson's proposition (see Proposition 5.3 in the paper mentioned above), this would imply that $U$ is isomorphic to $X/Y$, a contradiction.


