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Suppose $\{F_n\}$ is sequence of compact subsets of the half-plane $\mathbb H=(0,+\infty)\times(-\infty,+\infty)$ with positive area such that $F_n\subset F_{n+1}$ and $\cup_n F_n=\mathbb H$. Do there exist a sequence of positive numbers $\{a_n\}$ and positive constants $\delta,\delta',\delta''$ such that

$$[\frac{1}{a_n},a_n]\times[0,a_n]\subset F_n\subset[\frac{\delta}{a_n},\delta'a_n]\times[-\delta'' a_n,\delta''a_n]$$
for $n>n_0$?

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    works for very nice compact sets. This might be a wishful thinking but if turns out to be true it has a nice consequence for so-called Følner sequences. – 2017-01-09
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    Just a remark, positive area does not imply nonempty interior. Look up fat Cantor sets – 2017-01-09

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No. Let $c_n$ be a monotone increasing sequence. For convenience let $d_n = \frac{c_n}{c_n-1}$. Then set $$F_n = \bigg( [c_n^{-1},c_n]\times[-c_n, c_n] - [d_n^{-1},d_n]\times[-d_n, d_n] \bigg) \cup \{1\}\times [-c_n, c_n] $$

Each $F_n$ is closed, bounded, so compact. It looks like a large rectangle with a smaller rectangle punched out and a line strung across the removed smaller rectangle. (The line there means that the $F_n$ indeed converge to $\Bbb{H}$.) As $n$ increases, the large rectangle lets larger and the removed rectangle gets smaller.

Because the inner rectangle you propose has $(1,\varepsilon)$ in its interior for small $\varepsilon$, whereas $(1,\varepsilon)$ is on the boundary of $F_n$ for each $n$, containment can never hold.