 Solution

Embed the ring of integers into ${\mathbb{R}}^2$ by its two real embeddings. Its image is a full Euclidean lattice of some covolume $d>0$. Fix $C>d$ and consider the symmetric rectangle $|{\sigma }_1(\alpha )|\le T$, $|{\sigma }_2(\alpha )|\le C/T$. Its area is $4C>4d$, so the Minkowski convex body theorem gives a nonzero algebraic integer ${\alpha }_T$ in it. Its algebraic norm is a nonzero integer with $|N({\alpha }_T)|\le C$.

As $T\to \infty$, infinitely many distinct ${\alpha }_T$ must occur: a finite collection of nonzero elements has a positive minimum of $|{\sigma }_2(\alpha )|$, whereas here that coordinate tends to zero. The principal ideals $({\alpha }_T)$ have norm $|N({\alpha }_T)|\le C$. There are only finitely many ideals of bounded norm: they are subgroups of bounded index in a rank-two free abelian group, and integer Hermite normal forms give finitely many possibilities at each such index. Thus infinitely many of these elements generate one ideal. Ratios of two generators of the same ideal are units, because inclusion in both directions supplies inverse algebraic integers. Among infinitely many distinct generators choose two whose ratio is not $\pm 1$. This gives a non-trivial unit without assuming Dirichlet's theorem.