Nested intervals proofs 
a. Let $I_n\subset \mathbb{R}^n$ be the open cells given by
  $I_n=(0,1/n)\times \cdots\times (0,1/n)$. Show that these cells are
  nested but that they do not contain any common point. 
b. let $J_n \subset \mathbb{R}^n$ be the closed intervals given by
  $J_n=[n,+\infty)\times \cdots\times [n,+\infty)$. Show that these
  intervals are nested, but that they do not contain any common point.

I need to show for nested that $I_{n+1} \subset I_n$, for all $n$ correct? So, take $(x_1,\ldots,x_p) \in I_{n+1}$ Which means that every $x_i >0$ and $x_i< 1/(n+1)$, but I know that $1/(n+1) < 1/n$ so every $x_i$ it is also in $(0, 1/n)$, which means that is in the product $I_n$, correct? To show that there are no common points, I know that it is an open interval that when tend to infinity it will keep decreasing thus not having any common points; I am not sure if that is correct?
For b. I do not know how to do it for infinity. 
 A: (a) Your argument to show that $I_{n+1}\subseteq I_n$ is correct. To show that $\bigcap_{n\in\Bbb Z^+}I_n=\varnothing$, let $x=\langle x_1,\dots,x_p\rangle\in\Bbb R^p$ be arbitrary. If some $x_k\le 0$, then clearly $x\notin I_1$, so $x\notin\bigcap_{n\in\Bbb Z^+}I_n$. If $x_k>0$ for $k=1,\dots,p$, let $a=\min\{x_1,\dots,x_p\}$, and choose $n\in\Bbb Z^+$ so that $\frac1n\le a$; clearly $x\notin I_n$. (You could actually set $a=\max\{x_1,\dots,x_p\}$, since you need only get $\frac1n$ less than or equal to one of the coordinates of $x$.)
(b) This really isn’t any different. If $x=\langle x_1,\dots,x_p\rangle\in I_{n+1}$, then $x_k\ge n+1$ for $k=1,\dots,x_p$, then clearly clearly $x_k\ge n$ for $k=1,\dots,p$, so $x\in I_n$, and therefore $I_n\supseteq I_{n+1}$. For any $x=\langle x_1,\dots,x_p\rangle$ we can let $a=\min\{x_1,\dots,x_p\}$ and choose a positive integer $n>a$; can $x$ possibly be in $I_n$?
Added: The picture below shows $I_2$ and $I_3$ in the case $p=2$; $I_2$ is the unbounded diagonally shaded region stretching off to the upper right, and $I_3$ is the similar cross-hatched region, also stretching off to the upper right, contained within it.

A: I think you need to use a different letter for the dimension; I shall use $d$.  Choose $x\in\mathbb{R}^d$. Suppose $n > \max_{1\le k \le d} x_k$. Then $x\not\in J_n$. Hence $\cap_n J_n = \emptyset$.
