As Terry mentions in the comments, the reason for the $\sqrt{5}$ is that the limiting case, the golden ratio, forces it. There is a very neat explanation of all of this in the classic number theory book by Hardy and Wright, pages 209 to 212. I give a brief sketch of the ideas.

- Why $\phi$ is the worst case.

As Hardy and Wright put it, "from the point of view of rational approximation, the simplest numbers are the worst. The "simplest" of all irrationals, from this point of view, is the number $\phi$."

The reason for this is that if we consider the best approximation for a given $\alpha$,

$$\left|\alpha - \frac{p_n}{q_n} \right| = \frac{1}{q_nq'_{n+1}} < \frac{1}{a_{n+1}q^2_n}$$

it is best when $a_{n+1}$ is large. But in the case of $\phi$, every $a_{n+1}$ is as small as possible.

- Why it leads to $\sqrt{5}$

The idea is to simply see what happens when we approximate $\phi$. It roughly goes like this:

$$\left|\phi - \frac{p_n}{q_n} \right| = \frac{1}{q_nq'_{n+1}} \sim \frac{1}{q^2_n}\frac{1}{1+2\phi}=\frac{1}{q^2_n\sqrt{5}}$$

- $\sqrt{5}$ is best possible

This follows easily by contradiction. There are no infinitely many $p$, $q$ such that

$$\alpha=\frac{p}{q}+\frac{\delta}{q^2}$$ and $$|\delta|<\frac{1}{\sqrt{5}}$$

Now any proof of the theorem should look convincing enough, knowing where the $\sqrt{5}$ it presupposes comes from.

**EDIT**. I include for completeness a nice alternative proof brought up by Marty and Halbort in the comments.

L. R. Ford, "Fractions" (Amer. Math. Monthly, Vol 45, No 9 (Nov 1938))