I had to answer three questions about accumulation points. I think my work is correct, but I'd appreciate if someone would look over them for me. (I'm not sure if I read question 2 correctly.)
In my proofs, I will define $x$ as an accumulation point of $S \subseteq \mathbb{R}$ if the defining condition holds: $\forall \epsilon > 0, \exists y \in S$ s.t. $y \neq x$ and $y \in (x-\epsilon,x+\epsilon)$.
(1)
Find an infinite subset of $\mathbb{R}$ that does not have an accumulation point in $\mathbb{R}$.
Take $\mathbb{Z}$, an infinite subset of $\mathbb{R}$. We can show that $\nexists x \in \mathbb{R}$ to satisfy the defining condition. Suppose there were some $x$ s.t. $\forall \epsilon > 0, \exists y \in \mathbb{Z}$ s.t. $y \neq x$ and $y \in (x-\epsilon, x+\epsilon)$, iff $|y-x| < \epsilon$. Note that as $y \neq x$, $y-x \neq 0$, so $0<|y-x|<\epsilon$. Then let $\epsilon = \frac{|y-x|}{2}$, so $0 < \epsilon < |y-x|$. So we can conceive an $\epsilon$ s.t. $y \notin (x - \epsilon, x+\epsilon)$, so the defining condition doesn't hold.
(2)
Find a bounded subset of $\mathbb{R}$ that does not have an accumulation point in $\mathbb{R}$.
Take $X = \{1,2,3\}$, so $X \subseteq \mathbb{Z} \subseteq \mathbb{R}$. See (1).
(3)
Find all the accumulation points of $\mathbb{R} \backslash \mathbb{Q}$ in $\mathbb{R}$. Note that $\mathbb{R} \backslash \mathbb{Q} = \mathbb{I}$, the irrationals.
The rationals and the irrationals are both dense in $\mathbb{R}$, so it follows that both $(\mathbb{I} \cap (x-\epsilon,x+\epsilon))\backslash\{x\} \neq \emptyset$ and $(\mathbb{Q} \cap (x-\epsilon,x+\epsilon))\backslash\{x\} \neq \emptyset$ hold for any $\epsilon >0$. We know this because all rationals and irrationals are reals, so any such rational or irrational $\pm \epsilon$ is also a real, so there exists both a rational and an irrational between any two reals $x-\epsilon, x+\epsilon$. Therefore, no matter how small we make our $\epsilon$, we will still have both a rational and an irrational in $(x-\epsilon,x+\epsilon)$. As the reals are the union of the disjoint sets $\mathbb{Q},\mathbb{I}$, it must be that $\mathbb{R}$ is the set of accumulation points for the irrationals.