Reducing Exact Cover to Subset Sum
Basic idea: In the exact cover problem, each element is reduced to a number. Then, each set is reduced to the sum of the numbers it covers. Finally, set $k$ to be the sum of all numbers.

This reduction always works one way: If a solution exists for the exact cover problem, a solution also exists for the subset sum problem.

The hard part is to prove the other way. Trivial numbers don't work. Suppose the elements are reduced to $1,2,3,4$, and the sets are 


$S_1=\{3,4\}$
$S_2=\{3\}$


The reduction to subset sum is the numbers $7,3$ and $k=10$, which return YES but there's no solution to the exact cover problem!

We need to be smarter in choosing the numbers. Noticed the flaw in the previous numbers is that $3$ can replace $1,2$. We don't want any number to be replaced. Hence, we want the numbers to have sufficiently large gaps. Suppose in an exact cover problem of $n$ elements, the first element is reduced to $1$. How large should the second number be? There are $2^{n-1}$ sets that can include the first number. So, the second element should be reduced to a number larger than $2^{n-1}$ such that no matter how many times $1$ appears, it cannot replace the second number. Let's just reduce the second element to $2^n$. The same argument applies to the third argument, so we reduce it to $2^n\times 2^n=4^n$.

In conclusion, the $i$-th element in the exact cover problem has a value of $2^{in}$. Each set is reduced to the sum of numbers it covers.

Now we showed that exact cover is in NP-Hard. We need to prove that it is also in NP in order to show that it is NP-Complete. I think that's relatively simple. If you need help for that, I'll provide some hints for you.