The Baire Category Theorem states:
Theorem: Let $X$ be a complete metric space. Suppose that each element of the sequence $\{U_n\}$ of open sets of $X$ is dense. It holds that $\cap^\infty U_n$ is also dense in $X$.
Some definitions:
Density: A subset $Y$ of a metric space $X$ is dense if $\overline{Y} = X$.
Incorrect proof: Let $x \in X$. It suffices to show that $ B(x,\epsilon) \cap (\cap U_n) \neq \emptyset$. Suppose $B(x,\epsilon) \cap (\cap^\infty U_n) = \emptyset$, then, there exists some $n_0 \in \mathbb{N}$ such that $B(x,\epsilon) \cap U_{n_0} = \emptyset$ but this contradicts $\overline{U_{n_0}} = X$ and the theorem is proved. Q.E.D.
This proof ought to be incorrect since it doesn't make use of the completeness of $X$ but I can't really tell why this argument is incorrect. Any hint is appreciated. Thanks!
Thinking a bit more about it I have the following counter example:
Consider the intervals $U_1 = (-2,-1)$ and $U_2 = (1,2)$. Consider the open ball in $\mathbb{R}$, $B(0,2) = (-2,2)$. Clearly $B(0,2) \cap (U_1 \cap U_2) = \emptyset$ but it is not true that $B(0,2) \cap U_1 = \emptyset$ or $B(0,2) \cap U_2 = \emptyset$. So that implication is false.