 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 $L=\mathbb{Q}(i)$. Sincein $\mathbb{Z}[i]$, 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.