1. $(-\sqrt{2}, \sqrt{2}) \cap \mathbb{Q}$ is both closed and open on $\mathbb{Q}$. However, for a given space E, Rudin Theorem 2.27 states that if $X$ is a metric space and $E \subset X$ then $E = \bar{E}$ if and only if E is closed. [$\bar{E} = E \cup E'$, $E'$ denotes the set of all limit points of E in X]. How does this work with the idea that a sequence of rationals can converge to a number outside of the rationals?

Let $X = \mathbb{Q}$. $E = (-\sqrt{2}, \sqrt{2}) \cap \mathbb{Q} \subset \mathbb{Q}$. The set of all limit points E' must be in X, so it can only contain rationals.