For a random variable $X$, does $E[X^2]=0$ imply that $X=0$?
Does this work:
By contradiction. For each $X=x_i$ $$ E[X^2] = \sum_i x_i^2 \cdot p(x_i) $$ So $E[X^2] \ge x_i^2 p(x_i)$. Now if there exists an $x_i>0$, then $E[X^2] \ge x_i^2 p(x_i) > 0$ contradiction.
I'm confused since I see an argument here: $ E( |X| ) = 0 $ implies $X = 0$ except possibly on a null set that's a bit more complicated
One main nitpick: you need to assume there exists an $|x_i|>0$ with $P(X=x_i)=p_i>0$, as you have not ruled out your random variable could be negative, nor have you implied that $P(X=x_i)>0$.
For your last question, It depends how far along you are in your study of probability theory. From what you've written, you're currently studying discrete random variables, so for your purposes this is fine.
To be inline with the proof you linked to, you want to show $P(X= 0)=1$, e.g. $\{X\neq 0\}$ is a null set with respect to $P,X$. This is what you can accomplish by addressing the nitpick above.