Partitioning a convex object without cutting existing convex subsets $C$ 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:

 A: 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 $C$.
