I've heard that if a polynomial-time solution was found for an NP-hard problem, that would mean that all NP-hard problems can be solved in polynomial time. Is this true, and why is that?
NP-hard can be read as "as hard to solve as any problem in the class NP". So if we can solve one of these hard problems in polynomial time, then -- by the meaning of NP-hard -- we can solve all problems in the class NP in polynomial time.
The reason why your statement is not true as it is, is that there are NP-hard problems which are not inside the class NP. These problems are "much harder than" any problem in the class NP and so, being able to solve all problems in NP in polynomial time does not mean that these "harder than all of NP" problems can also be solved in polynomial time. Check the Wikipedia entry on NP-hard for an example of such a "harder-than NP, but also NP-hard" problem. https://en.wikipedia.org/wiki/NP-hard
A problem which is both (i) NP-hard, and (ii) in NP is said to be NP-complete. If a polynomial-time solution was found for an NP-hard problem, that would mean that all NP-complete problems can be solved in polynomial time (because they are all in NP).
[1] There is a fine point here turning on what "can be solved" means. If we take "can be solved" to mean "there exists an algorithm", then this statement is fine. Otherwise ... but that is an arcane point anyway.
Yes. It is because every problem in NP can be "reduced" (transformed into) an NP-complete problem in polynomial time. Therefore if a polynomial time solution for an NP-complete problem exists, the algorithm:
Have problem: 1) Transform problem into NP-complete problem (polynomial time) 2) Solve NP-complete problem (polynomial time)
Two polynomial time steps make a polynomial time algorithm.
The proof that every problem in NP is polynomial-time reducible to an NP-complete problem is pretty long and complicated, involving the formalism of Turing Machines, but it can be found in most textbooks on the theory of computation and I believe it is called the Cook-Levin Theorem (that proves the 3SAT is NP-complete, and then most other theorems show that 3SAT is reducible to them).