Given a (closed) separable subspace $M$ of $\ell\_\infty$, I am interested in conditions implying that the quotient $\ell\_\infty/M$ is isomorphic to a subspace of $\ell\_\infty$. 


It is not difficult to see that being $M$ reflexive is sufficient, and [Bourgain](http://www.ams.org/journals/proc/1980-078-02/S0002-9939-1980-0550499-2/S0002-9939-1980-0550499-2.pdf) proved that $\ell\_\infty/c\_0$ does not admit an equivalent strictly convex norm (while $\ell\_\infty$ does). So $\ell\_\infty/c\_0$ is not isomorphic to a subspace of $\ell\_\infty$. 



Quotients of $\ell_\infty$ by separable subspaces
While I do not have a complete answer to the OP’s question, I made enough observations that I think it is worthwhile to record them as an answer.


(1) If $X$ and $Y$ are isomorphic (closed) subspaces of $\ell\_\infty$, then $\ell\_\infty/X$ embeds into $\ell\_\infty$ iff $\ell\_\infty/Y$ embeds into $\ell\_\infty$. 


Indeed, it is clear that $\ell\_\infty/X$ embeds into $\ell\_\infty$ iff
$(\ell\_\infty \oplus \ell\_\infty)/(X \oplus \{0\})$ embeds into $\ell\_\infty$, so without loss of generality we can assume that $\ell\_\infty$ embeds into both $\ell\_\infty/X$ and into $\ell\_\infty/Y$. We then get from [LR, [Theorem 3(i)] that every isomorphism from $X$ onto $Y$ extends to an automorphism of $\ell\_\infty$.


(2) Suppose that $Y$ is a subspace of $\ell\_\infty$ and $X$ is a subspace of $Y$. Assume that $\ell\_\infty/Y$ embeds into $\ell\_\infty$. Then $\ell\_\infty/X$ embeds into $\ell\_\infty$ iff $Y/X$ embeds into $\ell\_\infty$. 


The “only if” part is clear because $Y/X$ embeds into $\ell\_\infty/X$. The other direction is an easy consequence of Lindentrauss’ observation that if $U\subset V$ and both $U$ and $V/U$ embed into $\ell\_\infty$, then $V$ embeds into $\ell\_\infty$. To prove this observation, let $T$ be an isomorphism from $U$ into $\ell\_\infty$ and $S$ an isomorphism from $V/U$ into $\ell\_\infty$. The space $\ell\_\infty$ is $1$-injective (this is immediate from the Hahn-Banach theorem), so $T$ extends to a bounded linear mapping (which we also denote by $T$) from $V$ into $\ell\_\infty$. Denoting the quotient map from $V$ to $V/U$ by $Q$, we see that $T \oplus SQ: x\mapsto (Tx, SQx)$ defines an isomorphism from $V$ into $\ell\_\infty \oplus\_\infty \ell\_\infty \equiv \ell\_\infty$.


(3) Suppose that $X$ is a subspace of of $\ell\_\infty$ and $X$ is isomorphic to $Y^\*$ for some separable $Y$. Then $\ell\_\infty/X$ embeds into $\ell\_\infty$.


Indeed, $Y$, being separable, is isomorphic to the quotient space $\ell\_1/W$ for some subspace $W$ of $\ell\_1$, and $W^\perp$ in $\ell\_\infty$ is isomorphic to $X$. But $\ell\_\infty/{W^\perp}$ is isometric to $W^\*$, which embeds into $\ell\_\infty$ because $W$, being separable, is a quotient of $\ell\_1$. Hence by (1), the space $\ell\_\infty/X$ also embeds into $\ell\_\infty$.


(4) Suppose that $X$ a subspace of of $\ell\_\infty$ and $X$ embeds into a separable conjugate space. Then $\ell\_\infty/X$ embeds into $\ell\_\infty$. 


This follows from (3), (2), and the fact that every separable space embeds into $\ell\_\infty$. 


(5) $\ell\_\infty/{c\_0}$ does not embed into $\ell\_\infty$. Hence if $X$ is a separable subspace of of $\ell\_\infty$ and $X$ contains a subspace isomorphic to $c\_0$, then $\ell\_\infty/X$ does not embed into $\ell\_\infty$. 


The second statement follows from the first and (1), (2). The first statement is another observation that I think is due to Lindenstrauss; namely, that $\ell\_\infty/{c\_0}$ contains an isometric copy of $c\_0(\Gamma)$ with $|\Gamma|=2^{\aleph\_0}$ (consider the image in $\ell\_\infty/{c\_0}$ of characteristic functions of $2^{\aleph\_0}$ infinite subsets of the natural numbers all of whose pairwise intersections are finite). The space $c\_0(\Gamma)$ does not embed into $\ell\_\infty$ because the unit vector basis for the space is a non separable set that has weakly compact closure, and every weakly compact subset of the dual to a separable space is separable. 


(6) If $X$ is a subspace of $\ell\_\infty$ that is isomorphic to $L\_1(0,1)$, then $\ell\_\infty/X$ embeds into $\ell\_\infty$. Hence if $X$ is a subspace of $\ell\_\infty$ that embeds into $L\_1(0,1)$, then $\ell\_\infty/X$ embeds into $\ell\_\infty$.


The second statement follows from the first by what is now standard reasoning. For the first statement, note that $L\_1(0,1)$ embeds into $C[0,1]^\*$ as a complemented subspace, and by (3), if $Y$ is a subspace of $\ell\_\infty$ isomorphic to $C[0,1]^\*$, then $\ell\_\infty/Y$ embeds into $\ell\_\infty$. Now use (2) and (1).


[LR] J. Lindenstrauss and H. P. Rosenthal, Automorphisms in $c\_0$, $\ell\_1$, and $m$, Israel J. Math. 7 (1969), 227–239.


