I do not know if the following qualifies as "short" or "elementary": but it does not follow the usual pattern through Minkowski's Convex Body Theorem. Rather, it mimics the classical proof of Mordell–Weil: honestly, it *copies* it, but I write the details down to give a direct feeling of what comes into the game.

First of all, Hermite—Minkowski tells you that $\mathcal{O}_K^\times/(\mathcal{O}_K^\times)^2$ is finite, as follows. For every unit $u$, the extension $K(\sqrt{u})/K$ has discriminant whose norm equals a *bounded* power of $2$, because every integral element $r+s\sqrt{u}\in \mathcal{O}_{K(\sqrt{u})}$ (with $r,s\in K$) has minimal polynomial $f(X)=X^2-2rX+r^2-us^2$ with $2r,r^2-us^2\in\mathcal{O}_K$. This shows that $\mathcal{O}_{K(\sqrt{u})}/\mathcal{O}_K(\sqrt{u})$ is a finite group killed by $4$ and of rank bounded by $2[K\colon\mathbb{Q}]$. The discriminant of $K(\sqrt{u})/K$ is then $2\sqrt{u}\cdot[\mathcal{O}_{K(\sqrt{u})}\colon\mathcal{O}_K(\sqrt{u})]$
and thus we get a bound $\operatorname{Disc}_{K(\sqrt{u})/\mathbb{Q})}\leq \operatorname{Disc}_{K/\mathbb{Q}}^2\cdot 2^{a(K)}$ for a constant $a(K)$ depending on $K$ only. It follows that the quotient $\mathcal{O}_K^\times/(\mathcal{O}_K^\times)^2$ classifies quadratic extensions of $K$ (this is easy: Kummer theory works over any field, in this case of characteristic different from $2$) of degree bounded by $2[K\colon\mathbb{Q}]$ and discriminant bounded by $\operatorname{Disc}_{K/\mathbb{Q}}^22^{a(K)}$, and this is a finite set of extensions by Hermite–Minkowski. One might say that Hermite–Minkowski is not trivial, which is true: but it is easier than the full proof of Dirichlet's Unit Theorem, in the sense that it comes well before it in almost all books I know, and Hermite's original proof consists of elementary, although quite tedious, algebra not involving any topology.

Now comes the ''Mordell–Weil" part: consider the usual height
$$
H(u)=\sqrt[{[K\colon\mathbb{Q}]}]{\prod_{\sigma\in\mathcal{M}_K^\infty}\max\{1,\vert\sigma(u)\vert\}^{\varepsilon(\sigma)}}\qquad\text{ for all }u\in\mathcal{O}_K^\times
$$
where the product is over all infinite places and $\varepsilon(\sigma)=1$ if $\sigma(K)\subseteq\mathbb{R}$, and $2$ otherwise. It is immediate to see that $H(u)\geq 1$ for all $u$, that $H(uv)\geq H(u)H(v)$ for all $u,v$ and that $H(u^{m})=H(u)^m$ for all $u$ and $m\geq 1$; and it is a classical result that the Northcott property holds, namely that for every given bound $B$ there are only finitely many elements in $\mathcal{O}_K^\times$ such that $H(u)\leq B$: this is elementary again, using simply that algebraic integers of bounded degree and bounded height are roots of polynomials in $\mathbb{Z}[X]$ with bounded coefficients and degree, hence a finite number of polynomials. A proof of this can be found on page 503 of Northcott's original paper

Northcott, D. (1949). An inequality in the theory of arithmetic on algebraic varieties. Mathematical Proceedings of the Cambridge Philosophical Society, 45(4), 502-509. doi:10.1017/S0305004100025202

By the first part, we know that there are finitely many units $\eta_1,\dots,\eta_r$ representing the elements in $\mathcal{O}_K^\times/(\mathcal{O}_K^\times)^2$; we also fix now a bound $B$ so that by the Northcott property there are only finitely many units $v_1,\dots,v_s$ of height bounded by $B$. Pick any $u\in\mathcal{O}_K^\times$: for every $n\geq 1$ we can write
$$
u=\eta_{i_0}u_1^2,\quad u_1=\eta_{i_1}u_2^2,\quad\dots \quad u_{n-1}=\eta_{i_{n-1}}u_{n}^2
$$
and thus
$$
u=\Big(\prod_{j=0}^n\eta_{i_j}^{2^j}\Big)u_{n+1}^{2^n}:=\eta(u;n)\cdot u_{n}^{2^n}.
$$
with $\eta(u;n)$ belonging to the subgroup generated by $\eta_1,\dots,\eta_r$; moreover, the above equation shows
$$
H(u)\geq H(u_n^{2^n})=H(u_n)^{2^n}\Rightarrow H(u_n)\leq H(u)^{1/2^n}\overset{n\to+\infty}{\longrightarrow} 1\quad(u\text{ is fixed}).
$$
Therefore, if we choose $n$ to be big enough, we get $u_n\in\{v_1,\dots,v_r\}$ and therefore $u$ belongs to the subgroup generated by $\{\eta_1,\dots,\eta_r,v_1,\dots,v_s\}$, showing that $\mathcal{O}_K^\times$ is finitely generated.

lookin any algebraic number theory book that proves the Unit Theorem? Property (i) is nearly always just the first part of the proof! That and the upper bound $r_1 + r_2 -1$ are comparatively easy. The only really hard part of the full Unit Theorem is showing the rank is $r_1 + r_2 -1 $. $\endgroup$Elementary and Analytic Theory of Algebraic Numbers(which is basically the same as the one in Ireland and Rosen's book suggested by Myshkin and Francesco Polizzi): After seeing Perdry's proof of Krull's intersection theorem, I convinced myself that there might have been published an equally slick proof of (i). After all, the bound implied by the proof in Narkiewicz's or Ireland and Rosen's book is stillverytight. Does this sound completely implausible? $\endgroup$Number Fields, pp. 142-143? $\endgroup$Number Theory. To my eyes, the proof is indeed neater than the ones I've seen so far, though this is mainly due to the way in which it's presented, not really to the circle of ideas on which it relies. In any case, it's getting definitely clear that my expectations were ungrounded: Since you've never seen an elementary proof of (i) that doesn't essentially yield, as part of the argument, the upper bound $r_1+r_2-1$ on the rank, it's highly probable that the kind of short and elementary proof I was looking for isn't (yet?) known. $\endgroup$