How to show it is contained in a convex hull? There are $(d+1)f$ points (denote the set of all points as $S$) in $\mathbb{R}^d$, that can be divide into $d+1$ disjoint sets $F_1,...,F_{d+1}$, each set of size $f$. If we have
$$
\mathcal{H}(F_i)\bigcap \mathcal{H}(S-F_i)=\emptyset
$$
for all $1\leq i\leq d+1$. How can we show the following:
$$
\mathcal{H}(S)-\bigcup_{i=1}^{d+1}\mathcal{H}(S-F_i)\subseteq \operatorname{conv}(v_1,...,v_{d+1})
$$
where $v_i\in F_i$, $1\leq i\leq d+1$
 A: $\def\conv{\mathop{\rm conv}}\let\emp\varnothing$This is the answer if @Douglas Zare is correct, so that  $\conv(\cdot)$ means the same as $\mathcal H(\{\cdot\})$. 
We omit the condition that $|F_i|$ are the same, thus considering any disjoint finite sets $F_i$ with $S=\bigcup_i F_i$. Notice that we may keep only those points of $F_i$ which are the vertices of $T=\conv S$; this just makes the claim sharper. Due to the compactness reason, we may assume that the vertices of the dual polyhedron of $\conv $ are in general position; thus every $d$ vectors connecting a vertex of $\conv S$ with other vertices are independent.
If $|F_i|=1$ for every $i$, the claim is trivial. So assume that $|F_1|>1$. In this case, since $F_1$ is separated from $F_j$ with $j>1$, there exists an edge $uv$ of $T$ with $u,v\in F_1$. This edge belongs to some $(d-2)$-dimensional face $P$ of $T$ which is a part of a $(d-1)$-dimensional face $Q$. 
Now rotate the hyperplane spanned by $Q$ around the affine subspace defined by $P$. Let $w$ be the first vertex we meet. Then the hyperplane $\alpha$ spanned by $P$ and $w$ separates some vertex of $Q$ from the rest vertices of $T$. Then $\alpha$ contains the vertices of at most $d-1$ of the $F_i$ (since it contains two vertices from $F_1$), so the whole simplex it cuts out of $T$ is contained in some $\conv (S-F_i)$. Now we may remove this vertex and proceed by induction.
