What's the relation between the dominating set and vertex cover?
Since there is no single minimal dominating set nor minimal vertex cover, I interpret your question as: 


  There exist one minimal dominating set $D$ and one minimal vertex cover $C$ such that $D \subseteq C$. 


No. 

If by any graph you also consider non-connected graphs, then you have a trivial counterexample in $G=(V,E)$ with $V\neq\emptyset$ and $E=\emptyset$, 
i.e. at least one vertex and no edges. 
The minimal vertex cover is $C=\emptyset$, since there are no edges to cover, 
but the only dominating set is $D=V$, and clearly $D\not\subseteq C$.

Even if you restrict it to connected graphs, the answer is still no. Consider the cycle graph $G=(V,E)$ of six vertexes $V=\{a,b,c,d,e,f\}$ and $E=\{(a,b), (b,c), (c,d), (d,e),(e,f),(a,f)\}$. 


The vertex cover sets of minimal size have three elements, obtained by skipping one vertex in two: $\{a,c,e\}$ and $\{b,d,f\}$. 
The dominating sets of minimal size are obtained with opposing vertexes, $\{a,d\}$, $\{b,e\}$, and $\{c,f\}$. 


None of the dominating sets is a subset of any vertex cover set.