[proofplan]
We show that every element $x \in K$ with $|x| > 1$ cannot satisfy any monic polynomial equation over $\mathcal{O}_K$, by using the strong triangle inequality to show that the leading term $x^n$ strictly dominates the lower-order terms. Therefore the only elements of $K$ that are integral over $\mathcal{O}_K$ lie in $\mathcal{O}_K$, proving $\mathcal{O}_K$ is integrally closed.
[/proofplan]
[step:Show that $|x| > 1$ implies $x$ satisfies no monic polynomial over $\mathcal{O}_K$]
Let $x \in K$ with $|x| > 1$, and let $a_0, a_1, \ldots, a_{n-1} \in \mathcal{O}_K$ (so $|a_i| \leq 1$ for each $i$). We show $x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \neq 0$.
Compute the absolute value of the lower-order terms using the strong triangle inequality. For each $i \in \{0, 1, \ldots, n-1\}$:
\begin{align*}
|a_i x^i| = |a_i| \cdot |x|^i \leq |x|^i \leq |x|^{n-1} < |x|^n.
\end{align*}
The first inequality uses $|a_i| \leq 1$, the second uses $i \leq n - 1$ and $|x| > 1$ (so $|x|^i$ is non-decreasing in $i$), and the strict inequality uses $|x| > 1$.
By the strong triangle inequality applied to the sum $a_{n-1} x^{n-1} + \cdots + a_0$:
\begin{align*}
|a_{n-1} x^{n-1} + \cdots + a_0| \leq \max_{0 \leq i \leq n-1} |a_i x^i| \leq |x|^{n-1} < |x|^n = |x^n|.
\end{align*}
Since $|a_{n-1} x^{n-1} + \cdots + a_0| < |x^n|$, the [Isosceles Triangle Principle](/theorems/???) (applied with the roles of $x^n$ and the remaining sum) gives
\begin{align*}
|x^n + a_{n-1} x^{n-1} + \cdots + a_0| = |x^n| = |x|^n > 0.
\end{align*}
In particular, $x^n + a_{n-1} x^{n-1} + \cdots + a_0 \neq 0$.
[guided]
The key idea is that in a non-archimedean field, when one term in a sum has strictly larger absolute value than all others, the sum has the same absolute value as the dominant term. This is the [Isosceles Triangle Principle](/theorems/???): if $|u| \neq |v|$, then $|u + v| = \max(|u|, |v|)$.
Here the dominant term is $x^n$: its absolute value $|x|^n$ strictly exceeds the absolute value of every other term $a_i x^i$ (since $|a_i| \leq 1$ and $|x| > 1$ force $|a_i x^i| \leq |x|^i \leq |x|^{n-1} < |x|^n$). Grouping all the non-leading terms into a single sum $s = a_{n-1}x^{n-1} + \cdots + a_0$, the strong triangle inequality gives $|s| \leq \max_i |a_i x^i| \leq |x|^{n-1} < |x|^n$. So $|x^n| > |s|$, and the Isosceles Triangle Principle yields $|x^n + s| = |x^n| > 0$.
This argument fails in the archimedean case: over $\mathbb{R}$, the polynomial $x^2 - 2$ has a root $\sqrt{2}$ with $|\sqrt{2}| > 1$, but $\sqrt{2}$ is integral over $\mathbb{Z}$. The non-archimedean property (strong triangle inequality) is essential.
[/guided]
[/step]
[step:Conclude that $\mathcal{O}_K$ is integrally closed in $K$]
We have shown: for every $x \in K$ with $|x| > 1$ and every monic polynomial $p(t) = t^n + a_{n-1}t^{n-1} + \cdots + a_0 \in \mathcal{O}_K[t]$, we have $p(x) \neq 0$. Equivalently, no element of $K \setminus \mathcal{O}_K$ is integral over $\mathcal{O}_K$ (since $K \setminus \mathcal{O}_K = \{x \in K : |x| > 1\}$).
Taking the contrapositive: every element of $K$ that is integral over $\mathcal{O}_K$ must satisfy $|x| \leq 1$, i.e., must lie in $\mathcal{O}_K$. This means the integral closure of $\mathcal{O}_K$ in $K$ is contained in $\mathcal{O}_K$. Since $\mathcal{O}_K$ is trivially contained in its own integral closure (every $x \in \mathcal{O}_K$ satisfies the monic polynomial $t - x \in \mathcal{O}_K[t]$), the integral closure equals $\mathcal{O}_K$.
[guided]
Let us carefully assemble the logical structure.
We have proved in the previous step: if $x \in K$ satisfies $|x| > 1$, then for every monic $p \in \mathcal{O}_K[t]$, $p(x) \neq 0$. In other words, no $x$ with $|x| > 1$ can be a root of any monic polynomial over $\mathcal{O}_K$.
Recall that $K \setminus \mathcal{O}_K = \{x \in K : |x| > 1\}$: by definition $\mathcal{O}_K = \{x \in K : |x| \leq 1\}$, so any element outside $\mathcal{O}_K$ must have absolute value strictly greater than $1$.
Therefore: no element of $K \setminus \mathcal{O}_K$ is integral over $\mathcal{O}_K$. Equivalently, every element of $K$ integral over $\mathcal{O}_K$ belongs to $\mathcal{O}_K$.
This gives the inclusion $\widetilde{\mathcal{O}_K} \subseteq \mathcal{O}_K$, where $\widetilde{\mathcal{O}_K}$ denotes the integral closure of $\mathcal{O}_K$ in $K$. For the reverse inclusion, every $x \in \mathcal{O}_K$ satisfies the monic polynomial $t - x \in \mathcal{O}_K[t]$ (since $x \in \mathcal{O}_K$), so $\mathcal{O}_K \subseteq \widetilde{\mathcal{O}_K}$. Together, $\widetilde{\mathcal{O}_K} = \mathcal{O}_K$, which is the definition of $\mathcal{O}_K$ being integrally closed in $K$.
[/guided]
[/step]
