Proof of proper subsets 
I know that the procedure for formulating this proof is to let x $\in$ $B$ be arbitrary and then I need to show that x $\in$ $A$. 
I've started with $2$ $<$ $x$ $\leq$ $3$ and separated it into two inequalities, $2$ $<$ $x$ and $x$ $\leq$ $3$, but I am stuck on how I can make these look like $x^2$$-$$9$ $\leq$ $0$ and $x^2$ $-$ $4$ $>$ $0$.
 A: Simple manipulations will give us inequalities like $A$:


*

*Since $2<x$, $x^2>4$ and $x^2-4>0$.

*Since $x\le 3$ and $x$ is constrained to be positive by $2<x$, $x^2\le9$ and $x^2-9\le0$.


Thus $x\in B\implies x\in A$ and $B\subseteq A$. To show properness of this inclusion ($A\subset B$), see that $x=-3$ is in $A$ but not $B$.
A: 

$$\left . 
\begin{array}{l}
x \in \color{green}B \implies 2 <x <=3 \implies x > 2 \implies x^2 > 4 \implies x^2-4>0\\
x \in \color{green}B \implies 2 <x <=3 \implies x \le 3 \implies x^2 \le 9 \implies x^2 -9 \le 0 
\end{array}
\right\}
\implies x \in \color{red}A$$
So $\color{green}B \subset \color{red}A$.
But the reverse inclusion $\left (\color{red}A  \subset \color{green}B\, \right)$ is not true, because for $x=-3:$ 
$$x \in \color{red}A,\quad x \notin \color{green}B$$
So $\color{green}B$ is a proper subset of $\color{red}A$.
A: $A$ is a proper subset of $B$ if $B \subseteq A$ and there exists an $x \in A$ such that $x \notin B$.
To show $B \subseteq A$: Let $x \in B$. Then, $x \leq 3 $ and $ x > 2$. Squaring both sides and moving everything to the left we have $x^2 - 9 \leq 0$ and $x^2 -2>0$, so $x \in A$. Thus $B \subseteq A$. 
Note that $(-3)^2 - 9 \leq 0$ and $(-3)^2 - 4 > 0$, so $-3 \in A$, but clearly $-3 \notin B$.
Putting these together, we can see that $A$ is a proper subset of $B$
A: Set $A=${$x \in \mathbb{R} | x^2 \le 9$ and $x^2>4$}.
1) $x^2 \le 9$ , i.e. $-3 \le x \le 3$.
2) $x^2 >4$, i.e. $x <-2$ or $x >2$.
1) and 2) : $x \in [-3,-2)$ or $x \in (2,3]$.
Rewriting:
$A=${$x| x \in [-3,-2)$ or $x \in (2,3]$}$.
$B=(2,3]$, definition.
Hence $B \subset A$ (Why proper?) 
