To execute strategy $(1),$ first sort the list in time $n^2$ and then remove the first $k$ elements, which takes time $k.$ So the total time taken is $n^2 + k.$

For strategy $(2),$ to remove the $k$ smallest elements, we need to scan through the list, look for the smallest element and remove it, then repeat this process $k$ times. For a list of length $i,$ finding and removing the smallest element takes time $i$, as we perform $i − 1$ comparisons to find it then $1$ removal. So the overall time taken for strategy $(2)$ is $n+(n − 1)+\cdots+(n − k + 1) = nk − \frac{k(k − 1)}{2}.$

To prove that strategy $(2)$ is faster, we must show that $nk − \frac{k(k − 1)}{2} < n^2 + k$ for all $n$ and all $k.$ We know that $1 \leq k \leq n$ so an immediate upper bound for the left-hand side is $nk − \frac{k(k − 1)}{2} \le n^2 − \frac{k(k − 1)}{2},$ which is indeed smaller than $n^2 + k.$