Question: For every Banach space $X$ and its subspace $Y$, is there a complemented subspace $Z$ in $X$ such that $Y \subset Z \subset X $ and $\operatorname{card}(Y)=\operatorname{card}(Z)$ i.e., $Y$ and $Z$ have the same cardinality?
The answer is yes if $\operatorname{card}(X)=\mathfrak c$ (i.e., Continuum), the proof is easy by taking a hyperplane $H$ such that $Y \subset H \subset X$, then $\operatorname{card}(Y)=\operatorname{card}(H)=\mathfrak c$.
But I don't know whether it holds for general spaces.
The above paper says that there is a Banach space (of continuous functions on a totally disconnected compact Hausdorff space) of density κ bigger than the continuum which has no infinite-dimensional complemented subspaces of density continuum or smaller. In particular no separable infinite-dimensional subspace has a complemented superspace of density continuum or smaller. Hence, for the space C(K) constructed by P.~Koszmider, for each separable subspace Y of C(K), there exist no complemented subspace Z containing Y such that card(Z)=card(Y), in fact, such space Z has card lager than c.