Is it possible construct a bounded linear non-compact operator $T:E \longrightarrow F$ such that maps from weakly convergent sequences of $E$ into strongly convergent sequences in $F$?
This is not possible if $E$ and $F$ are Banach spaces, and $E$ is reflexive. The requested property is called complete continuity. Completely continuous operators are compact:
Every bounded sequence $(x_n)$ has a weakly converging subsequence, hence $(Tx_n)$ has a strongly converging subsequence. This shows that bounded sets are mapped by $T$ to sequentially relatively compact sets, which are relatively compact sets in the Banach space. That is, $T$ is compact.
Note that this does not apply to $l^1$, which has the Schur property that weak and strong convergence coincides.