proof by contradiction: a composite $c$ has a nontrivial factor $\le \sqrt c$ 
Let $c$ be a positive integer that is not prime. Show that there is some positive integer $b$ such that $b \mid c$ and $b \leq \sqrt{c}$.

I know this can be proved by contradiction, but I'm not sure how to approach it. Usually I write the proof in the form $P \rightarrow Q$, and then if we can prove $P \land \neg Q$ is false, $P \rightarrow Q$ must be true.
In this case, I wrote it as:
If $c$ is a composite, positive integer, then $b \mid c$ and $b \leq \sqrt{c}$, for some positive integer $b$.
I'm guessing that as long as I assume that $b \nmid c$ or $b > \sqrt{c}$, then this is still valid as $\neg Q$; that is, I don't have to assume the converse of both parts of $Q$?
Moving on, if $b > \sqrt{c}$, and $b \mid c$, then $br=c$ for some integer $r$, which means $r < \sqrt{c}$.
And this is where I get stuck.

I: That there exists some $b$ with $b|c$ is true by the definition (c is not prime).
Now we assume that - as II - there is no $b \leq \sqrt c$.
Since there has to be some $b$ due to I, this $b$ now has to meet $b > \sqrt c$.
Now $z = \frac{c}{b}$ surely is an integer that divides $c$ too, and, furthermore, $z \leq \sqrt c$. Therefore, by assuming II, we disprove II - Contradiction, q.e.d.
