Can this proof be valid for $\bigcap_{n \in \mathbb{N}}(-1/n,1/n) =\{0\}$?

Question

Assume $∀x, x \in \{0 \}$. Thus $x=0$. So $∀n, x∈(-1/n, 1/n)$ since $0 \in (-1/n, 1/n)$. Hence $$∀n, \{0\} \subset \bigcap_{n \in \mathbb{N}}(-1/n,1/n).$$ Assume $ x \in \bigcap_{n \in ℕ}(-1/n,1/n)$. So $-1/n ≤ x ≤ 1/n$. By Squeezing Theorem, lim $x = 0$ as the limit for $-1/n$ and $1/n$ equals $0$ when $n→∞$. Hence $x=0$ since $x$ is continuous on $ℝ$. So $x \in \{0\}$. Thus $\bigcap_{n \in ℕ}(-1/n,1/n) \subset \{0\}$. Therefore $$\bigcap_{n \in ℕ}(-1/n,1/n) = \{0\}$$

Answer 1

The Squeezing Theorem should not be necessary. We can prove the nontrivial direction by contradiction.

Choose any $x \in \bigcap_{n \in \mathbb N}(-1/n,1/n)$. Suppose, towards a contradiction, that $x \neq 0$. Then by the Archimedean Property, we know that there is some $k \in \mathbb N$ such that $k > 1/|x| \iff |x| > 1/k$. But recall that: $$ x \in \bigcap_{n \in \mathbb N}\left(\frac{-1}{n}, \frac{1}{n} \right) \subseteq \left(\frac{-1}{k}, \frac{1}{k} \right) $$ It follows that $|x| < 1/k$, a contradiction. So $x = 0$, as desired. $~~\blacksquare$