Before this writing the smallest order for which the nonexistence of simple groups of that order is not explicitly demonstrated on this site is 480; this self-answered question aims to fill that gap. (See Prove there are no simple groups of even order $<500$ except orders $2$, $60$, $168$, and $360$., the sole answer of which describes an outline for proving the titular statement but does not describe how to establish the claim for order $480$. In particular, this question-answer pair completes that outline in the sense that all cases not explicitly resolved therein are covered by other, linked answers on this site.) Other answers are, of course, welcome.
How does one show that there are no simple groups of order $480$?
Suppose $G$ is a simple group of order $480 = 2^5 \cdot 3 \cdot 5$.
Since $G$ is simple, if it has a proper subgroup of index $k$, then $G$ embeds into $S_k$; as $480 \not\mid |S_k|$ for $k \leq 7$, any proper subgroup of $G$ has index $\geq 8$.$^*$ Sylow's Third Theorem then gives that the count $n_2 := |\operatorname{Syl}_2(G)|$ satisfies $n_2 \mid 15$ and $n_2 = |P : N_G(P)| \geq 8$ (for any, equivalently, every $P \in \operatorname{Syl}_2(G)$), leaving only the possibility$$n_2 = 15 .$$
Now consider the conjugation action of $P \in \operatorname{Syl}_2(G)$ on $\operatorname{Syl}_2(G)$. Since $N_P(Q) = P \cap Q$ (see, e.g., this answer), and the size of the $P$-orbit of any $Q \in \operatorname{Syl}_2(G)$ is $|P \cdot Q| = |P : N_P(Q)| = |P : P \cap Q|$. So, the action has $1$ fixed point ($P$ itself), and it partitions the set $\operatorname{Syl}_2(G) \setminus \{P\}$ of the remaining $14$ Sylow $2$-subgroups into orbits whose sizes are factors of $|P| = 2^5$ larger than $1$.
Since $2^2 \not\mid 14$, there is at least $1$ orbit of length $2$, hence some $R \in \operatorname{Syl}_2(G)$ such that $|P : P \cap R| = 2$; denote $T := P \cap R$. In particular, $T$ has index $2$ in both $P$ and $R$ and hence is normal in both. Since $N_G(T)$ contains both $P$ and $R$, $|N_G(T)| \geq 2^5 \cdot 3 = 96$, and hence $[G : N_G(T)] \leq \frac{480}{96} = 5$. Now, either $N_G(T) = G$, so that $T \trianglelefteq G$, which contradicts the simplicity of $G$, or $N_G(T)$ is a proper subgroup of index $\leq 5$, which contradicts our observation that every proper subgroup of $G$ must have index $\geq 8$.
$\mathbf{{}^*}$Remark In fact, if $G$ embeds in $S_k$, it must embed in $A_k$; otherwise $G \cap A_k$ would be an index-$2$---hence normal---subgroup of $G$, a contradiction. Moreover, Conway et al.'s ATLAS of Simple Groups shows that $A_8$ has no subgroups of order $480$, so in fact any proper subgroup of $G$ would have to have index $\geq 10$.