Prove that if $\displaystyle a_n $ is a sequence such that every subsequence of $\displaystyle a_n $ has a further subsequence converging to $\displaystyle 0 $, then $\displaystyle a_n \to 0 $.

Let $\displaystyle \sigma: \mathbb{N} \to \mathbb{N} $ be a strictly increasing function. Suppose that $\displaystyle b_{n} = a_{\sigma(n)} $ is a subsequence of $\displaystyle a_n $. Then $\displaystyle c_{n} = b_{\sigma(n)} \to 0 $ is a subsequence of $\displaystyle b_n $. If $\displaystyle \varepsilon > 0 $, then there is an $\displaystyle N \in \mathbb{N} $ such that $\displaystyle n \geq N $ implies that $\displaystyle c_{n} \in (-\varepsilon, \varepsilon) $.

From here, I have to show that $\displaystyle a_n \in (-\varepsilon, \varepsilon) $.

Now what? $\displaystyle c_n \leq b_n \leq a_n $. Could I use the Squeeze theorem?