Proof checking : Show that $\bigcap\limits_{n=1}^{\infty}\left(-\frac{1}{n},\frac{1}{n}\right)= \{0\}. $ Can anyone check my proof please? Thank you.

Show that $$\bigcap\limits_{n=1}^{\infty}\left(-\frac{1}{n},\frac{1}{n}\right)= \{0\}.$$


Let take an arbitrary $x \in \textbf{R}$ such that $|x|\leq 1/n$  for all $n$ in $\textbf{N}$.
Then $$ -\frac{1}{n}\leq x\leq\frac{1}{n}.$$
We can now say that $$x \in \bigcap\limits_{n=1}^{\infty}\left(-\frac{1}{n},\frac{1}{n}\right) .$$
We will now show that $x=0$.
We have: $$ -\frac{1}{n}\leq x \leq\frac{1}{n}.$$
Let's now take the limit $n\to\infty$:
$$\lim_{n\to\infty}-\frac{1}{n} \leq x \leq \lim_{n\to\infty}\frac{1}{n}.$$
$$0\leq x \leq0.$$
By the BigMac theorem, we conclude that $x=0$.
 A: $0 \in (-\frac{1}{n}, \frac{1}{n})$ for all $n$, and so $0 \in \bigcap_{n \in \mathbb{N}}(-\frac{1}{n}, \frac{1}{n})$
Suppose for contradiction that there exists some $x \in \bigcap_{n \in \mathbb{N}}(-\frac{1}{n}, \frac{1}{n})$ so that $x \neq 0$. Wlog, suppose $x>0$. Use the archimedean property to show that there exists $N \in \mathbb{N}$ so that $\frac{1}{N}<x$. Then show that for $n>N$, we have a contradiction.
There is a slight error in your proof, as has been noted in the comment section. Additionally, when you say that $|x|<\frac{1}{n}$ for all $n$, how are you justifying this? Really, you're letting $x=0$ anyway, why not begin the proof by showing the intersection to be nonempty?
A: 
Let's now take the limit $n\to\infty$:$$lim_{n\to\infty}-\frac{1}{n}<x< lim_{n\to\infty}\frac{1}{n}.$$

This part is incorrect and should be 
$$
\lim_{n\to\infty}-\frac{1}{n}\leq x\leq \lim_{n\to\infty}\frac{1}{n}$$
which implies that
$$
0\leq x\leq 0
$$
and thus $x=0$. 

Also, note that $0<x<0$ means $0<x$ and $x<0$, which is a false statement whatever $x$ is. I have only seen "Big Mac Theorem" before as a stronger statement of the so called "Hamburger's Theorem".
