= Solution
Choose the real embeddings $\sigma_i$ and one from each complex pair $\tau_j$. The <logarithmic embedding of number field units> is
$$
\lambda(u)=(\log|\sigma_1u|,\ldots,\log|\sigma_ru|,2\log|\tau_1u|,\ldots,2\log|\tau_su|).
$$
It is a <group homomorphism> from multiplication to addition. A unit has absolute <norm> one, so the coordinate sum is zero and its image lies in a Euclidean hyperplane of dimension $r+s-1$.
Here is the required discreteness argument. There are only finitely many algebraic integers of $K$ whose conjugate absolute values are bounded by a fixed constant: their monic minimal <polynomials> have degree at most $n$, and each integer coefficient is an elementary symmetric function of bounded roots, hence belongs to a fixed finite range. There are finitely many such <polynomials> and finitely many roots of each. A bounded set of logarithms bounds all conjugate absolute values, so every compact region in the logarithmic hyperplane meets the unit image in finitely many points. Thus the image is a discrete additive <subgroup>.
Use the allowed Euclidean-group theorem: a discrete <subgroup> of a finite-dimensional real <vector space> is free abelian on finitely many vectors, with <rank> at most the dimension. Therefore $\lambda(\mathcal O_K^\times)$ has <rank> at most $r+s-1$. Its <kernel> is finite by the same bounded-conjugate argument, now with every absolute value one. Each element of a finite multiplicative group is a <root of unity>, and conversely every <root of unity> has all logarithms zero. Thus the <kernel> is exactly the <finite group> of <roots of unity>, and exactly the <torsion subgroup> of units.
Lift generators of the free logarithmic image to units; together with the finite <kernel> they generate every unit. Hence
$$
\boxed{\mathcal O_K^\times\text{ is finitely generated abelian of rank at most }r+s-1,\quad
\operatorname{tors}(\mathcal O_K^\times)=\mu(K).}
$$
This proves the requested upper bound without invoking the existence half of the <Dirichlet unit theorem>. The field $\mathbb Q(\sqrt{11})$ is a subfield of $\mathbb R$, and its only real <roots of unity> are $\boxed{1,-1}$.
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