Hurwitz's Theorem: Let $\zeta$ be an irrational number, then there exist infinitely many coprime integers $p$ and $q$ such that
$$\lvert \zeta - \frac{p}{q} \rvert < \frac{1}{\sqrt{5} q^2}$$
What I want to show is that if we substitute the $\sqrt5$ above for any larger real number, then there would be a counterexample to the inequality; namely, $\varphi = \frac{1 + \sqrt5}{2}$.
To do so, we first explore some inequalities relating $\varphi$ and $\frac{p}{q}$.
Lemma 1: Let $\frac{p}{q} \in Q$, then $\lvert \varphi - \frac{p}{q} \rvert \lvert \bar{\varphi} - \frac{p}{q} \rvert \geq \frac{1}{q^2}$, where $\bar{\varphi} = \frac{1-\sqrt5}{2}$ is the conjugate of $\varphi$.
Lemma 2: Let $\frac{p}{q} \in Q$, and $m \in R$. If $\lvert \varphi - \frac{p}{q} \rvert < \frac{1}{mq^2}$, then $\lvert \bar{\varphi} - \frac{p}{q} \rvert < \frac{1}{mq^2} + \sqrt5$.
Suppose now that there are infinitely many coprime integers $p_i$ and $q_i$ such that $\lvert \varphi - \frac{p_i}{q_i} \rvert < \frac{1}{mq^2_i}$, then we can use the lemmas to arrive at an upper bound for m:
$$\frac{1}{q_i^2} \leq\lvert \varphi - \frac{p_i}{q_i} \rvert\lvert \bar{\varphi} - \frac{p_i}{q_i} \rvert < \frac{1 + mq_i^2\sqrt5}{m^2q_i^4}$$
$$\frac{1}{q_i^2} < \frac{1 + mq_i^2\sqrt5}{m^2q_i^4}$$
$$q_i^2m^2 < 1 + mq_i^2\sqrt5$$
$$q_i^2m^2 - \sqrt5q_i^2m -1 < 0$$
$$q_i^2(m^2-\sqrt5m) - 1 < 0$$
Recall that we want to find an upper bound for $m$ regardless of $q_i$. Thus, we have to consider $\lim \limits_{i \to \infty} q_i^2(m^2-\sqrt5m) - 1$. Note that if $q_i \to \infty$ as $i \to \infty$, then the limit would be negative if and only if $m^2 - \sqrt5m$ is not positive, which happens only when $m \leq \sqrt5$. This would suffice to show that $\sqrt5$ is the greatest constant for which Hurwitz's Theorem is true. Indeed, since there are finitely many numerators for each denominators, the sequence $q_1, q_2, ...$ must tend to infinity.
Appendix
Proof of Lemma 1: We consider three cases. First, that $\frac{p}{q} > \varphi > \bar{\varphi}$: we carefully remove the absolute values to get $(\frac{p}{q} - \varphi)(\frac{p}{q} - \bar{\varphi})$, which simplifies to $\frac{p^2}{q^2} -\frac{p}{q}(\varphi + \bar{\varphi}) + \varphi\bar{\varphi}$. Note that $\varphi + \bar{\varphi} = 1$ and $\varphi\bar{\varphi} = -1$, so we end up with $\frac{1}{q^2}(p^2 - pq - q^2)$.
Second, that $\varphi > \frac{p}{q} > \bar{\varphi}$: similarly, we remove the absolute values to get $(\varphi - \frac{p}{q})(\frac{p}{q} - \bar{\varphi})$, which simplifies to $-\frac{p^2}{q^2} +\frac{p}{q}(\varphi + \bar{\varphi}) - \varphi\bar{\varphi} = \frac{1}{q^2}(-p^2 + pq + q^2)$.
And third, that $\varphi > \bar{\varphi} > \frac{p}{q}$: we get $(\varphi - \frac{p}{q})(\bar{\varphi} - \frac{p}{q}) = \frac{p^2}{q^2} -\frac{p}{q}(\varphi + \bar{\varphi}) + \varphi\bar{\varphi} = \frac{1}{q^2}(p^2 - pq - q^2)$.
Observe that, in all three cases, we are left with the product of two things that must be positive. Thus $\lvert \varphi - \frac{p}{q} \rvert \lvert \bar{\varphi} - \frac{p}{q} \rvert = \frac{1}{q^2} \lvert p^2 -pq - q^2 \rvert$. If we consider $p^2 -pq - q^2$ as a polynomial on $p$, we quickly find out that it has no real roots, and thus, that $\lvert p^2 -pq - q^2 \rvert \geq 1$. Hence, $\lvert \varphi - \frac{p}{q} \rvert \lvert \bar{\varphi} - \frac{p}{q} \rvert \geq \frac{1}{q^2}$.
Proof of Lemma 2: We consider the same three cases as the first lemma.
$\frac{p}{q} > \varphi > \bar{\varphi}$: $\lvert \varphi - \frac{p}{q} \rvert = \frac{p}{q} - \varphi < \frac{1}{mq^2}$. Adding $\sqrt5$ to the inequality, we get $\frac{p}{q} - \bar{\varphi} < \frac{1}{mq^2} + \sqrt5$. Since $\frac{p}{q} - \bar{\varphi} > 0$, we can put the absolute value back on: $\lvert \frac{p}{q} - \bar{\varphi} \rvert = \lvert \bar{\varphi} - \frac{p}{q} \rvert < \frac{1}{mq^2} + \sqrt5$
$\varphi > \frac{p}{q} > \bar{\varphi}$: we use the following reverse triangle inequalities: $\lvert \varphi - \frac{p}{q} \rvert \geq \lvert \varphi \rvert - \lvert \frac{p}{q} \rvert$ and $\lvert \frac{p}{q} - \varphi \rvert \geq \lvert \frac{p}{q} \rvert - \lvert \varphi \rvert$. To remove the absolute values, we must denote whether $\frac{p}{q}$ is positive or negative.
$\frac{p}{q} \geq 0$: we use the latter inequality to arrive at $\lvert \frac{p}{q} \rvert - \lvert \varphi \rvert = -\varphi + \frac{p}{q} < \frac{1}{mq^2}$. In adding $\sqrt5$, we get $-\bar{\varphi} + \frac{p}{q} = \lvert -\bar{\varphi} + \frac{p}{q} \rvert = \lvert \bar{\varphi} - \frac{p}{q}\rvert < \frac{1}{mq^2} + \sqrt5$
$\frac{p}{q} < 0$: we now use the former inequality to arrive at $\lvert \varphi \rvert - \lvert \frac{p}{q} \rvert = \varphi + \frac{p}{q} < \frac{1}{mq^2}$. Note that $\bar{\varphi} < {\varphi}$, so we have that $\bar{\varphi} + \frac{p}{q} < \frac{1}{mq^2}$. If we add $\sqrt5$, we return to $\varphi + \frac{p}{q} < \frac{1}{mq^2} + \sqrt5$. One simply has to note that $0 < - \bar{\varphi} + \frac{p}{q} < \varphi + \frac{p}{q} < \frac{1}{mq^2} + \sqrt5$. Thus $\lvert -\bar{\varphi} + \frac{p}{q} \rvert = \lvert \bar{\varphi} - \frac{p}{q} \rvert$.
$\varphi > \bar{\varphi} > \frac{p}{q}$: we have that $\bar{\varphi} - \frac{a}{b} < \varphi - \frac{p}{q} < \frac{1}{mq^2}$. Hence, $\bar{\varphi} - \frac{p}{q} < \frac{1}{mq^2}$ becomes $\varphi - \frac{p}{q} < \frac{1}{mq^2} + \sqrt5$. $\bar{\varphi} - \frac{p}{q} < \frac{1}{mq^2} + \sqrt5$. Finally, $\lvert \bar{\varphi} - \frac{p}{q} \rvert < \frac{1}{mq^2} + \sqrt5$.