Proving or disproving properties of a set in $\mathbb{R}^2$ Consider a set $\Omega = \{(x_1, x_2) \in \mathbb{R}^2 \text{ s.t. } |x_2| \leq 1\}$. 
Is this set open? No. Consider a point $(x_1, 1)$. Then there is no way to construct an open ball with a nonzero radius at this point, which is the requirement for an open set. 
Is this set closed? Yes. Points of the form $(x_1, \pm1)$ are boundary points, and are included in the set (I don't know how to show this more formally)
Is this set bounded? No, because $\lim_{x_1 \rightarrow \infty} ||(x_1, x_2)||_2$ does not have an upper bound. $\forall x \in \Omega$, $||x|| \leq C$ is not satisfied for some $C \in (0, +\infty)$. 
Would appreciate any corrections on what I have written above, as well as advice in rigorous techniques for proving open/closed/bounded properties of sets. 
 A: I assume you are using the norm $\|(x,y)\|=\sqrt {x^2+y^2}\;.$ In particular 
 that $\|(x,y)\|\geq |y|.$
Re:Open. You should say there is no open ball $B$ with $(x_1,1)\in  B\subset \Omega.$ And you can prove it as follows: If $r>0$ and $B=B(p,r)$ is  the open ball centered at the point $p,$ with radius $r,$ let $s=\|p-(x_1,1)\|.$ Then $s<r$ so we have $$\|p- (x_1, 1+(r-s)/2)\|\leq \|p-(x_1,1)\|+\|(x_1,1)-(x_1,1+(r-s)/2)\|=$$ $$=s+(r-s)/2=(r+s)/2<r .$$ So $(x_1,1+(r-s)/2)\in B\setminus \Omega.$
Re: Closed. If $(b_n)_{n\in \Bbb N}$ is a convergent real sequence and $|b_n|\leq 1$ for all $n\in \Bbb N$ then $|\lim_{n\to \infty}b_n|\leq 1.$  If $((x_{1,n},x_{2,n}))_{n\in \Bbb N}$ is a sequence of members of $\Omega$ converging to $(x,y)$ then $(x_{2,n})_{n\in \Bbb N}$ converges to $y$. (Because $|x_{2,n}-y|\leq \|(x_{1,n},x_{2,n}-(x,y)\|,$ and $\|(x_{1,n},x_{2,n})-(x,y)\|\to 0$ as $n\to \infty).$  So $|y|\leq 1,$ so $(x,y)\in \Omega.$
Another way to show $\Omega $ is closed is to show that $\Bbb R^2 \setminus \Omega$ is open. If $(x,y)\in \Bbb R^2\setminus \Omega,$ let $r=|y|-1.$ No member of the open ball $B=B((x,y),r/2)$ lies in $\Omega.$ Because if $(x',y')\in B$ then $r/2>\| (x',y')-(x,y)\|\geq |y'-y|.$ So we have $$|y'|+r/2>|y'|+ |y-y'|\geq |y'+(y-y')|=|y|=1+r$$ implying $|y'|>1+r/2>1.$
Re:Unbounded. You haven't proved it .You have only re-stated it.  For any $x\geq 0 $ we have $(x+1,0)\in \Omega$ and $\|(x+1,0)\|=x+1>x$ so $x$ cannot be an upper bound for $\{\|p\|: p\in \Omega\}.$ 
