Inclusion-Exclusion Principle for basic combinatorics problem... How many ways are there to pick five people for a committee if there are six
  (different) men and eight (different) women and the selection must include at
  least one man and one woman?


I know to solve this with very basic combinatorial analysis (though I know the principle is also basic). We find the total people we can choose for the committee and proceed with cases: 1 where we have all men on the committee, 1 where we have all women on the committee and then we subtract the two cases from our total. This gives us
$${14 \choose 5} - {8 \choose 5}-{6 \choose 5}.$$

How would I compute for Inclusion-Exclusion Principle, though? I've no idea where to begin with it other drawing a venn diagram and labeling set M and W...
You have more-or-less done it already.

For $i \in \{0,1,\ldots\}$, let $a_i$ be the number of committees with $\geq i$ absent genders.

We include $a_0=\binom{14}{5}$.

We exclude $a_1=\binom{8}{5}+\binom{6}{5}$.

We include $a_2=0$.

We exclude $a_3=0$.

And so on.
In my opinion, this is the answer:

1m and 4 w : 6 X 8C4
2m and 3 w : 6C2 X 8C3
3m and 2 w : 6C3 X 8C2
4m and 1 w : 6C4 X 8

Since these are all mutually exclusive, they can all be added.