We know that if $X$ is a separable Banach space, then for every infinite dimensional Banach space $Y$, there exists an injective compact operator from $X$ to $Y$.


My query is for every Banach space $X$ (need not be separable ) do there exist a Banach space $Y$ and an injective compact operator $T:X\to Y$? 



Existence of injective compact operators
No, for cardinality reasons. The range of a compact operator is norm-separable hence has cardinality continuum (if non-zero). It is then enough to take $X$ to have bigger cardinality, for example, $X = \ell\_\infty^\*$. Then you have no chance of building such operators.


Another possibility for counterexamples comes with non-separable reflexive spaces (or, more generally, WCG spaces), which contain non-separable weakly compact sets. Injective bounded linear operators are then homeomorphic embeddings of such sets with respect to the weak topology, so cannot be compact.


