Suppose I wish to show that my decision problem $Q$ is NP-Hard. Why do I need to reduce from one problem $Q'$ of known hardness ?
Consider for instance the following situation:
Here, I have my set NP of problems, where problems $A$ and $B$ were proved to be NP-Complete through direct proof, and problems $C$, $D$, $E$, and $Q'$ were proved to be NP-Complete through reduction. I reduce from $Q'$ to $Q$, then later show $Q$ is in $P$. I agree that I then would be able to claim that problems $Q'$ and $B'$ can be claimed to also be in $P$, since we can just reduce them to $Q$. However, why can we make the same claims about $C$, $D$ and $E$ ? there is no (known) way of reducing from them to $Q$, we merely know that they may be "at least as hard" as the unlabeled problem and problem $A$, but can't it be possible that they are harder?
This is why intuitively, it seems to me like we should be showing ways to reduce from all "leaf" problems in NP, to show that $Q$ is also NP-Hard, even if I am told that this is not needed.