Open sets in the subspace topology 
Consider the set $Y=[-1,1]$ as a subspace of $\mathbb{R}$, where $\mathbb{R}$ is equipped with the standard topology. Which of the following sets are open in $Y$? Open in $\mathbb{R}$?
$A=\lbrace x \in \mathbb{R}:\frac{1}{2}<|x|<1 \rbrace$
$B=\lbrace x \in \mathbb{R}:\frac{1}{2}<|x|\leq1 \rbrace$
$C=\lbrace x \in \mathbb{R}:\frac{1}{2}\leq|x|<1 \rbrace$
$D=\lbrace x \in \mathbb{R}:\frac{1}{2}\leq|x|\leq1 \rbrace$
$E=\lbrace x \in \mathbb{R}:0<|x|<1$ and $\frac{1}{x} \notin \mathbb{N}$ $\rbrace$

I think I have figured out $A,B,C$, and $D$, but just looking for some help proving some of them rigorously. I don't know where to start with $E$.
Firstly, we rewrite $A=(-1,\frac{-1}{2}) \cup (\frac{1}{2},1)$, $B=[-1,\frac{-1}{2}) \cup (\frac{1}{2},1]$, $C=(-1,\frac{-1}{2}] \cup [\frac{1}{2},1)$, and $D=[-1,\frac{-1}{2}] \cup [\frac{1}{2},1]$.
We know sets are open in the subspace topology if they can be expressed as the intersection of $Y$ and some open set of $\mathbb{R}$. $A\subset Y$, so $A \cap Y=A$; also, $A$ is a union of basis elements of $\mathbb{R}$, so it is open in both $Y$ and $\mathbb{R}$.
If we let $U=(-2,\frac{-1}{2}) \cup (\frac{1}{2},2)$, then $B=U \cap Y$, and clearly $U$ is open in $\mathbb{R}$, so $B$ is open in $Y$ (but not in $\mathbb{R}$; it is a union of half-open, half-closed intervals).
For $C$, I am reasonably certain that it is not open in $Y$, nor in $\mathbb{R}$, but I'm not sure how to express it formally. The problem is that we can't use the same trick we did for $B$, because we need our intersection to be closed on the $\frac{-1}{2}$ and $\frac{1}{2}$ ends, but these are not endpoints of $Y$.
Same story for $D$ - it isn't open in either space.
 A: You can check all of them by checking the complements.
$Y \setminus A = [-\frac{1}{2},\frac{1}{2}] \cup \{-1\} \cup \{1\}$ which is closed.
$Y \setminus B = [-\frac{1}{2},\frac{1}{2}]$ which is closed.
$Y \setminus C = (-\frac{1}{2},\frac{1}{2}) \cup \{-1\} \cup \{1\}$ which is NOT closed.
$Y \setminus D = (-\frac{1}{2},\frac{1}{2})$ which is NOT closed.
$Y \setminus E = \{\frac{1}{x} \mid x\in \Bbb Z^*\} \cup \{0\}$ which is closed since it contains all of its limit point (i.e. $0$). (Sorry for the mistake made before)
As for using definition directly, $E = \{\bigcup (\frac{1}{x},\frac{1}{x+1}) \mid x \in \Bbb Z^+) \} \cup \{\bigcup (\frac{1}{x-1},\frac{1}{x}) \mid x \in \Bbb Z^-) \}$, which is a union of open set, so it is open. Apply subspace topology, it is still open in $Y$.
A: Checking the complements is a good method. But you can directly know which one is open or not by a useful theorem:
First, given an ordered set $X$, let us say that a subset $Y$ of $X$ is convex in $X$ if for each pair of points $a<b$ of $Y$, the entire interval $(a, b)$ of points of $X$ lies in $Y$. 
Theorem: Let $X$ be an ordered set in the order topology; let $Y$ be a subset of $X$ that is convex in $X$. Then the order topology on $Y$ is the same as the topology $Y$ inherits as a subspace of $X$. 
Now, since $Y=[-1,1]$ is convex in $\mathbb{R}$, the subspace topology of $Y$ is just the order topology of $Y$ whose base is the family of all intervals of forms $[-1,a),(a,b)$ and $(a,1]$. For example, $B=[-1,-\frac{1}{2})\bigcup(\frac 12,1]$ is open in $Y$ but is not open in $\mathbb{R}$.
