$C$ is a convex object in the plane. $D_1,\dots,D_n$ are pairwise-disjoint convex subsets of $C$ such that $D_1\cup\dots\cup D_n \subsetneq C$, like this:
Is it always possible to partition $C$ to $n$ convex subsets, $E_1,\dots,E_n$, such that $E_1\cup\dots\cup E_n = C$, and for every $i=1,\dots,n$: $D_i \subseteq E_i$? like this:
EDIT: following a hint by @dxiv, here is my solution.
By the Hyperplane separation theorem, for every pair $i\neq j$ there is a line $L_{ij}$ separating $D_i$ from $D_j$. The line $L_{ij}$ divides the plane to two half-planes: one of them (call it $H_{ij}$) contains $D_i$ and the other (call it $H_{ji}$) contains $D_j$.
For every $i$, define:
$$ E_i = C \cap \left( \cap_{j\neq i} H_{ij} \right) $$
Every $E_i$ is convex since it is an intersection of convex objects.
For every $i$, $D_i\subseteq E_i$ since it is contained in $C$ and in all the half-planes in the intersection.
The $E_i$ are pairwise-disjoint since for every $i\neq j$, the half-plane $H_{ij}$ contains $E_i$ and is disjoint from $E_j$.
Problem is, the union of the $E_i$ is not equal to $C$! As seen in the illustration below (where I kept only $D_1,D_2,D_4$), there is a blue triangle which is not contained in any of the $E_i$-s:
I think you are about one step away from constructing a counterexample.
Consider your last figure. The interiors of regions $E_1$, $E_2$, and $E_4$ are pairwise disjoint convex sets whose union is a proper subset of $C$.
The only convex subsets of $C$ that contain $E_1$ and do not intersect either $E_2$ or $E_4$ are subsets of the closure of $E_1$. That is, the convex subset of $C$ must of course contain $E_1$, and it may contain some or all of the points on the boundary of $E_1$, but it cannot contain any points "above" line $L_{24}$ or to the "right" of line $L_{12}$, because inevitably there would be line segments connecting those points to points of $E_1$ while passing through either $E_2$ or $E_4$.
If you want the $D_i$ to be closed sets, then let $D_i$ be the set of all points inside the $E_i$ shown in the figure, and at least $\epsilon$ away from the boundary of $E_i$, setting $\epsilon$ small enough so that there is not enough "wiggle room" between the $D_i$ to construct convex sets containing them that look much different from the $E_i$ in the figure. In particular you can ensure that most of the area of the blue triangle continues to be impossible to include in the desired partition of $C$.