In the aforementioned book in p. 12 right under the approximation lemma there is a corollary stating that:
"A Dedeking domain $A$ with only finitely many prime ideals is principal".
I can find different proofs elsewhere but I am really interested in the way that it can be proved with what Serre has already proved. The proof goes like this:
"It is enough to show that every prime ideal is principal. Let $p$ be one, there exists an $x$ in $A$ with $v_p(x)=1$ and $v_q(x)=0$ for $q\neq p$. Hence $xA=p$."
Obviously the problem is how can such an $x$ be found, since the approx. lemma only gives inequalities.
The statement of the approximation lemma can be found on a similar question, here. I would like to mention that up to now, Serre hasn't mentioned the so-called "Strict triangle inequality", neither the fact that in the approximation lemma the inequalities can be taken to be equalities.
Let $\mathfrak p_1, ... , \mathfrak p_n$ be all the primes of $A$. We want to show that $\mathfrak p = \mathfrak p_1$ is principal. Choose an $a \in \mathfrak p$ which is not in $\mathfrak p^2$. By the approximation theorem as stated by Serre, there exists an $x$ in the field of fractions of $A$ such that $$\nu_{\mathfrak p_i}(x-1) \geq 1$$ for $2 \leq i \leq n$, and
$$\nu_{\mathfrak p}(x-a) \geq 2$$
From here you can see that $x$ must lie in $A$.
For each $2 \leq i \leq n$, we have $x-1 \in \mathfrak p_i$, so $x \not\in \mathfrak p_i$. We also have $x - a \in \mathfrak p^2$. Since $\mathfrak p^2 \subseteq \mathfrak p$, and $a \in \mathfrak p$, we also have $x \in \mathfrak p$. But $x$ is not in $\mathfrak p^2$, since otherwise $a$ would be in $\mathfrak p^2$.
The conditions $x \not\in \mathfrak p_2, ... , \mathfrak p_n$, and $x \in \mathfrak p - \mathfrak p^2$ are equivalent to Serre's claim about $x$.