If $\liminf z_n = 0$, then there is a null sequence $(y_n)$ such that $\sum y_n = \infty$ and $\liminf y_n/z_n = 0$ 
Let $(z_n)$ be a sequence in $(0, \infty)$ with $\liminf z_n = 0$. Show that there are null sequence $(y_n)$ in $(0,\infty)$ such that $\sum y_n = \infty$ and $\liminf y_n/z_n = 0$.

There is a hint told me to construct $(y_n)$ as follows: choose a convergent subsequence $(z_{n_k}) \to 0$. Set
$$
y_{n_k} = z_{n_k}^2 \text{ for all } k \in \mathbb{N}, \quad y_n = 1/n \text{ otherwise}
$$
$(y_k)$ clearly is null and safisfies $\liminf y_n/z_n = 0$. My question is, is $(z_{n_k}) \to 0$ enough to conclude $\sum y_n = \infty$? Since there are instances that can make the sequence converges under this construction (e.g. $z_k = k^{-2}$ and pick subsequence $(z_{k})$ where $k \neq 2^n$ for any $n$), how do we ensure the existence?

You're right, the hint seems incomplete. You might try this: Start with the subsequence $z_{n_k}\to 0.$ We then divide $n_k$ into the two subsequences $n_{2k}$ and $n_{2k+1}.$ For $n\in \{n_{2k}\},$ do as you did and set $y_{n_{2k}}= z_{n_{2k}}^2.$ For $n\in \{n_{2k+1}\},$ set $y_{n_{2k+1}}= 1/k.$ For all other $n,$ if any, set $y_n=1/n.$
