 Solution

Suppose there were only finitely many prime numbers congruent to $5$ modulo $6$, and list them as $p_1,\dots ,p_k$. The list is nonempty since it contains $5$. Let $N=6p_1\cdots p_k-1$. It is greater than one and divisible by neither $2$ nor $3$. Every prime factor therefore has residue $1$ or $-1$ modulo $6$. Since $N\equiv -1(\mathrm{mod}6)$, at least one prime factor $q$ must have residue $-1$. But $N\equiv -1(\mathrm{mod}{p}_i)$ for every listed prime, so $q$ is not on the list. This contradiction proves there are infinitely many primes congruent to $5$ modulo $6$.