I was watching this Mathologer video (https://youtu.be/YuIIjLr6vUA?t=1652) and he says at 27:32
First, suppose that our initial chunk is part of a parabola, or if you like a cubic, or any polynomial. If I then tell you that my mystery function is a polynomial, there's always going to be exactly one polynomial that continues our initial chunk. In other words, a polynomial is completely determined by any part of it. [...] Again, just relax if all this seems a little bit too much.
So he didn't give a proof of the theorem in bold text – I think this is very important.
I understand that there always exists a polynomial of degree $n$ that passes through a set of $n+1$ points (i.e. there are finitely many custom points to be passed by, the chunk has to be discrete, like $(1,1),(2,2),(3,3),(4,5)$). But there also exists some polynomial of degree $m$ ($m\ne n$) that passes through the same set of points.
But how do I prove that there exists one and only one polynomial that passes through a set of infinitely many points?