Question: Let $R = k[x, y]$ be the polynomial ring in two variables and $I$ an ideal of $R$ such that $R/I$ is finite dimensional. I wonder if the following statement is correct? 

If $\mathfrak{m}$ is a maximal ideal in $R/I$, then its annihilator in $R/I$ is nonzero. 

Here, I assume $k$ is algebraically closed. Thanks!

Answer: 
Got it! As pointed out by @Wojowu, the set has to be countably infinite. So (i) wouldn't work because of the Bolzano-Weierstrass Theorem. Since $A$ is countable, its elements can be listed as $a\_1,...$, and in that list there exists a convergent subsequence. The limit of this sequence would be a limit point.
The first cannot exist as $[0,1]$ is limit point compact. The second obviously exists as $[0,1]\cap\mathbb{Q}$ shows. And for every space with a countable base we also have a countable base for any subset, so no uncountable discrete subset can exist.