Is destructuring a heap (taking down a heap) also O(n) like building a heap? If so, can the selection problem be solved by this method in O(n) time?
Delete-max is $O(\log n)$, so you have $O(n) + k O(\log n)$ which is $O(n)$ for fixed $k$. But if $k$ is not fixed, it can be up to $\frac{n}{2}$ (if it's greater, just use a min-heap instead of a max-heap), and  $O(n) + \frac{n}{2} O(\log n) = O(n \log n)$.

In fact, if there was some tricky way to do what you want, you would solve sorting in $O(n)$, not just selection, which is known to be impossible.