No, I really do; however, I have had a lot of homework in my other classes

So this is what your saying:

Let $\varnothing\not=B\subset(\mathbb N\cup A)$. \begin{align}B&=B\cap(\mathbb N\cup A)\\ &=(B\cap A)\cup (\mathbb N\cap B).\end{align} Let $a$ be the least element of the finite set $B\cap A$, if it exists. Let $m$ be the least element of $B\cap\mathbb N$, if it exists. By ordering principle of $\mathbb N$, then the min of the two element set $\{a,m\}$ is the least element of B.