How to prove Exact cover problem is NP Complete using Vertex Cover problem?
You can reduce from vertex cover in cubic graphs since it is known to be $\textrm{NP}$-hard. Let $G=(V,E)$ be the graph in the instance of vertex cover and, for $v \in V$, call $E_v = \{ (v,u) \mid (v,u) \in E \}$ the set of edges incident to $v$ in $G$.

Construct an instance of the exact cover problem as follows:


The elements to be covered are those in $E$. 
There are $7$ sets $S_1^v, \dots, S_7^v$ for each vertex $v \in V$, namely those in $2^{E_v} \setminus \{ \emptyset \}$.


If there is an exact cover $C$ for this instance, then there is a vertex cover of $G$ of size at most $|C|$, namely $\{ v \in V \mid \{S_1^v, \dots, S_7^v\} \cap C \neq \emptyset\}$.

If there is a vertex cover $C$ of $G$, then there is an exact cover of size at most $|C|$. Arbitrarily assign each edge $(u,v)$ of $G$ to one of the vertices in $\{u,v\} \cap C$. For a vertex $v$, let $A_v$ be the set of edges assigned to $v$ and notice that if $A_v \neq \emptyset$ then $A_v = S_i^v$ for some $i \in \{1,\dots, 7\}$. The sought exact cover is then $\{ A_v \mid v \in V \wedge A_v \neq \emptyset \}$.

To conclude the proof you just need to observe that exact cover belongs to $\textrm{NP}$ since the cover itself is a certificate that can be checked in polynomial-time.