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. 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 of $G_\epsilon$ 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.

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. 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.