 Solution

Assume first that $K$ is complete. If $x$ is integral over ${\mathcal{O}}_K$, its monic equation and the ultrametric inequality imply $|x{|}_L\le 1$, so $x\in {\mathcal{O}}_L$. Conversely, if $|x{|}_L\le 1$, part (i) gives $|\sigma (x){|}_L\le 1$ for every $K$-embedding $\sigma$. The coefficients of the minimal polynomial of $x$ are elementary symmetric polynomials in its conjugates, so they all lie in ${\mathcal{O}}_K$. Thus $x$ is integral over ${\mathcal{O}}_K$, proving the integral closure in a finite extension of a complete discretely valued field identity ${\mathcal{O}}_L={\overline{{\mathcal{O}}_K}}^L$.

Completeness is necessary. Give $\mathbb{Q}$ its $5$-adic absolute value, take $L=\mathbb{Q}(i)$, and choose the extension corresponding to the prime $(2+i)$ above $5$. Thenhas nonnegative valuation at $(2+i)$, so it belongs to the chosen valuation ring ${\mathcal{O}}_L$. At the conjugate prime $(2-i)$ it has negative valuation, so it does not lie in the integral closure of ${\mathbb{Z}}_{(5)}$ in $\mathbb{Q}(i)$. Hence the chosen valuation ring can be strictly larger than the integral closure when the base field is not complete.