Step 1: Know what optical isomerism needs.
A complex shows optical isomerism when it is chiral, meaning its mirror image cannot be placed on top of itself. Bidentate ligands like ethylenediamine ($en$) often make this possible.
Step 2: Check $[Co(NH_3)_4Cl_2]^+$.
This has only simple $NH_3$ and $Cl$ ligands. Its cis and trans forms both match their mirror images, so it is not optically active.
Step 3: Check $[Co(en)_2Cl_2]^+$.
This has two $en$ rings plus two $Cl$. The cis form of this complex has no plane of symmetry, so its mirror image cannot be superimposed. It is chiral and shows optical isomerism.
Step 4: Check $[Co(NH_3)_5Cl]^{2+}$.
With five $NH_3$ and one $Cl$, the complex has a plane of symmetry, so it is not optically active.
Step 5: Check $[Co(NH_3)_6]^{3+}$.
Six identical $NH_3$ groups make a fully symmetric shape, so it has many planes of symmetry and is not chiral.
Step 6: Pick the chiral one.
Only $[Co(en)_2Cl_2]^+$ in its cis form lacks a plane of symmetry, so only it shows optical isomerism.
Step 7: State the final answer.
The complex showing optical isomerism is:
\[ \boxed{[Co(en)_2Cl_2]^+} \]