= Solution
Assume first that $K$ is complete. If $x$ is integral over $\mathcal O_K$, its monic equation and the <ultrametric inequality> imply $|x|_L\leq1$, so $x\in\mathcal O_L$. Conversely, if $|x|_L\leq1$, part (i) gives $|\sigma(x)|_L\leq1$ 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 $5$. Then
$$
x=\frac{2+i}{2-i}
$$
has 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.
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