I have to prove that if $\langle u,v \rangle = 0$ for all $v$, then $u = 0$.
I thought about this:
$\langle u,v \rangle = 0$
$\langle u,v \rangle = \langle 0,v \rangle$
$ u = 0 $
I'm almost sure that this is very wrong, but it was the only way that I could use to prove it.
I think the problem is a little weird as well, because $u=0$ to any $v$. However, if $v = 0$, then $u$ doesn't have to be equals to zero necessarily.
You need to get your quantifiers right, the way you wrote it in the question is wrong. The correct thing is if $\langle u,v\rangle =0$ for all $v$, then $u=0$. The "for all $v$" comes before the conclusion, as it is an assumption. That said, the proof is very simple: if $\langle u,v\rangle=0$ for all $v$, you're allowed to choose $v=u$. That $\langle u,u\rangle = 0$ implies $u=0$ is one of the axioms in the definition of inner product, so we are done.