TRUE or FALSE: If a sequence converges to $0$, then it has a summable subsequence Let $\left( {{a_n}} \right)_{n = 1}^\infty $ be a sequence such that $\mathop {\lim }\limits_{n \to \infty } |{a_n}| = 0$. Prove that there are is subsequence ${a_{{n_k}}}$ so that $\sum\limits_{k = 1}^\infty  {{a_{{n_k}}}} $ converges.
Proof: Let $\left( {{a_n}} \right)_{n = 1}^\infty $ be a sequence and $\mathop {\lim }\limits_{n \to \infty } |{a_n}| = 0$ and ${a_{{n_k}}}$ be a subsequence.
Since $\lim\limits_{n \to \infty } |{a_n}| = 0$, let $\varepsilon  > 0$ , there is $N > 0$ so that 
    $$||a_n| - 0| < \frac{\varepsilon }{{m - i}},$$ that is $$|a_n| < \frac{\varepsilon }{{m - i}}$$
for all $n \geqslant N$ and $m,i \in N$ which $m > i.$
Consider subsequence ${a_{{n_k}}}$ , $k = i,i + 1,...,m - 1,m,[m - i{\text{ terms}}]$
Since $|{a_n}| < \frac{\varepsilon }{{m - i}}$ hence $$\begin{align}|{a_{{n_k}}}| &< \frac{\varepsilon }{{m - i}}|{a_{{n_{i + 1}}}} + {a_{{n_{i + 2}}}} + ... + {a_{{n_{m - 1}}}} + {a_{{n_m}}}|\\ &\leqslant |{a_{{n_{i + 1}}}}| + |{a_{{n_{i + 2}}}}| + ... + |{a_{{n_{m - 1}}}}| + |{a_{{n_m}}}|\\ &< \frac{\varepsilon }{{m - i}} + \frac{\varepsilon }{{m - i}} + ... + \frac{\varepsilon }{{m - i}}\\ &= \frac{{(m - i)\varepsilon }}{{m - i}}\\ &= \varepsilon\end{align}$$
$$\left|\sum_{k = i + 1}^m a_{n_k}\right|  < \varepsilon $$
By CAUCHY CRITERION FOR SERIES $\sum\limits_{k = 1}^\infty  {{a_{{n_k}}}} $ converges.

If I am not mistaken, then your purported proof shows that every subsequence is summable, hence if it was correct it would show that every sequence converging to zero is summable. This is not the case.
A correct proof will start with a given sequence converging to zero and then show how to produce, using properties of that particular sequence, a subsequence which is summable, i.e. one which not only converges to zero but converges fast enough.
