= Solution
Choose a $K$-basis of $L$. Any extension $|\cdot|_L$ defines a norm on the finite-dimensional $K$-vector space $L$, and all norms on such a space over a complete valued field induce the same topology. Hence any two extensions induce the same topology and are <equivalent absolute values>. The coordinate sup norm is complete because $K$ is complete, so the equivalent topology defined by $|\cdot|_L$ is complete as well. This proves the <unique extension of an absolute value to a finite extension> and the completeness of $L$.
Completeness is essential. Give $K=\mathbb Q$ its $5$-adic value and take $L=\mathbb Q(i)$. Since
$$
5=(2+i)(2-i)
$$
in $\mathbb Z[i]$, the two primes $(2+i)$ and $(2-i)$ define two inequivalent extensions. For example, $(2+i)/(2-i)$ has positive valuation for one and negative valuation for the other. This is the <valuation ring need not equal the integral closure over a noncomplete field> example.
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