An elementary, short proof that the group of units of the ring of integers of a number field is finitely generated Dirichlet's unit theorem states that (i) the group of units, $\mathscr{U}_K$, of the ring of integers of a number field $K$ is finitely generated, and (ii) the rank of $\mathscr{U}_K$ is equal to $r_1 + r_2 - 1$, where $r_1$ is the number of real embeddings and $r_2$ the number of conjugate pairs of complex embeddings of $K$. 

Q. What about a reference to an elementary, short proof of (i)?

In principle, (i) is much weaker than (ii), so the question shouldn't be so implausible. The reason why I'm asking is that I seem to have an alternative,  extravagant proof of a well-known result, but for the approach to be of any potential interest I need (a reference to) an elementary, short proof of (i).
 A: This part of Dirichlet's unit theorem is the only one needed in the standard proof of the Weil-Mordell theorem over number fields, so you probably can find it in a few introductory books on elliptic curves.
In particular, "A Classical Introduction to Modern Number Theory" by Kenneth Ireland and Michael Ira Rosen contains a full proof in pages 326-327 (they call it "the weak Dirichlet unit theorem").
A: 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.
A: This follows, via the standard "logarithmic embedding", from the fact that a discrete subgroup of the additive group $\mathbb{R}^n$ is a lattice.
An elementary proof of (i) can be found in the book 
K. Ireland, M. I. Rosen: A Classical Introduction to Modern Number Theory,
see in particular p. 326, $\S 3$: "The weak Dirichlet unit theorem".
A: William Stein gives a short argument (Section 8.1).
A: $O_K$ is a discrete cocompact subring in $K=\mathbf{R}^{r_1}\times\mathbf{C}^{r_2}$. So the group $U_K^\times$ of invertible elements of $O_K$ is a discrete subgroup in $K^\times$, the group of invertible elements of $K$, which is isomorphic to $\mathbf{R}^{r_1+r_2}\times W$ with $W$ compact (isomorphic to $(\mathbf{Z}/2\mathbf{Z})^{r_1}\times(\mathbf{R}/\mathbf{Z})^{r_2}$). 
A discrete subgroup of such a Lie group is finitely generated and its $\mathbf{Q}$-rank is bounded above by $r_1+r_2$. Indeed the composite homomorphism $U_K\to \mathbf{R}^{r_1+r_2}$ is proper and hence maps injectively $U_K$ to a discrete subgroup of $\mathbf{R}^{r_1+r_2}$.
(Probably the upper bound by $r_1+r_2-1$ comes from an additional argument saying that $U_K$ goes into some subgroup of elements of norm one, but the point here is precisely that we don't need this to get the above weaker upper bound $r_1+r_2$.)
