Sorry, my comment yesterday was rushed and this will be only slightly less so. Thus it is more of a (misleading?) hint. It is perhaps easier to view $E$ etc. as a $C^\infty$ bundle equipped with a $\bar\partial$ operator. Then the exact sequence $$0\to \mathcal{O}_X\to E\to \mathcal{O}_X\to 0$$ admits a $C^\infty$ splitting, with respect to which $$\bar \partial_E= \begin{pmatrix} \bar \partial&\alpha\\\ 0 &\bar \partial\end{pmatrix}$$ where $\alpha$ is a $(0,1)$-form. Integrability should force this to be a closed form. Factoring out the dependence on the splitting should give the Dolbeault class. A $C^\infty$ integrable connection on $E$, compatible with the sequence, can expressed similarly with $\alpha$ now just a closed $1$-form. So you see the inclusion naturally from this perspective.

Sorry, my comment yesterday was rushed and this will be only slightly less so. Thus it is more of a hint. It is perhaps easier to view $E$ etc. as a $C^\infty$ bundle equipped with a $\bar\partial$ operator. Then the exact sequence $$0\to \mathcal{O}_X\to E\to \mathcal{O}_X\to 0$$ admits a $C^\infty$ splitting, with respect to which $$\bar \partial_E= \begin{pmatrix} \bar \partial&\alpha\\\ 0 &\bar \partial\end{pmatrix}$$ where $\alpha$ is a $(0,1)$-form. Integrability should force this to be a closed form. Factoring out the dependence on the splitting should give the Dolbeault class. A $C^\infty$ integrable connection on $E$, compatible with the sequence, can expressed similarly with $\bar \partial$ replaced by $d$ and $\alpha$ now just a closed $1$-form. So you see the inclusion naturally from this perspective.