I just stumbled upon the 4-year-old question. Let me add another application to the list.
Claim. Let $A$ be a commutative Noetherian ring. If primes $\mathfrak p, \mathfrak r$ satisfy $\mathfrak p < \mathfrak r$, then the interval $(\mathfrak p, \mathfrak r)$ contains infinitely many primes or zero primes.
Chat. Here I am working in the ordered set of primes, so $\mathfrak p<\mathfrak r$ means $\mathfrak p\subsetneq \mathfrak r$. Here $(\mathfrak p, \mathfrak r)$ denotes the set of primes strictly intermediate to $\mathfrak p$ and $\mathfrak r$. Finally, this question about the structure of intervals in the ordered set of primes is asked in order to understand Spec($A$).
Reasoning for Claim. Suppose otherwise that $(\mathfrak p, \mathfrak r)$ is finite and nonempty. Shrink this interval to a minimal such one, and we end up with an interval containing some $\mathfrak q\neq \mathfrak p, \mathfrak r$ such that $\mathfrak p\prec \mathfrak q\prec \mathfrak r$ and $(\mathfrak p,\mathfrak r)$ is finite. (The notation $\mathfrak p\prec \mathfrak q$ indicates covering, which means $(\mathfrak p,\mathfrak q)$ is empty.) When we have shrunk to a minimal such interval, no two distinct primes strictly between $\mathfrak p$ and $\mathfrak r$ will be comparable.
Factor by $\mathfrak p$ to reduce to the case $\mathfrak p=(0)$. Then localize at $\mathfrak r$ to reduce to the case where $\mathfrak r$ is maximal. Now we have $(0)\prec \mathfrak q\prec \mathfrak r$ and that $\mathfrak r$ is maximal. In this reduced case we are trying to show that it is impossible for an integral domain of Krull dimension $2$ to have only finitely many primes. This is where Prime Avoidance comes in. Assume that $\mathfrak q_1, \ldots, \mathfrak q_n$ is a complete list of the primes strictly between $(0)$ and $\mathfrak r$ (i.e., height-$1$ primes). By Prime Avoidance, it is impossible to have $\mathfrak r$ contained in $\cup_{i=1}^n \mathfrak q_i$, so choose $a\in \mathfrak r - \cup_{i=1}^n \mathfrak q_i$. By the Krull Principal Ideal Theorem, any prime minimal over $(a)$ must have height $1$, so must be one of the $\mathfrak q_i$'s. But we specifically chose $a$ so that it belongs to none of them, so we are done. $\Box$
The Noetherianness hypothesis is used to allow us to invoke the Krull Principal Ideal Theorem.