Proving a Certain Planar Measure Is Zero on Horizontal Lines 
Question: Suppose $\mu$ is a measure on $\mathbb{R}^{2}$ with respect to which all open squares are measurable. Suppose $\mu$ has the following
  property: there exists a constant $\alpha\geq 1$ such that if $Q$ and $Q'$ are
  any two open squares which are translates of one another and $\overline{Q}\cap\overline{Q}'\neq\emptyset$, then
$$\mu(\overline{Q})\leq\alpha\mu(Q'),$$
where $\overline{Q}$ denotes the closure of $Q$. Then horizontal lines have zero measure with respect $\mu$.

This is an old qual problem which I cannot solve. I know that by continuity of measure, it suffices to show that any horizontal line segment has $\mu$-measure zero. I tried imitating the proof that line segments have Lebesgue measure zero, but that approach quickly failed.
 A: By $\sigma$-finiteness of $\mathbb{R}^{2}$ and continuity of measure, it suffices to show that any open horizontal line segment $(a,b)\times\left\{c\right\}\subset\mathbb{R}^{2}$ has $\mu$-measure zero. Without loss of generality, we may assume that $c=0$. We will show that 
$$\lim_{n\rightarrow\infty}\mu\left((a,b)\times[0,\left|a-b\right|/n]\right)=0 \qquad (n\in\mathbb{N})$$
Then by monotonicity of measure, $\mu((a,b)\times\left\{0\right\})=0$.
For each $n\in\mathbb{N}$, we can write the $E_{n}:=(a,b)\times (0,\left|a-b\right|/n)$ as $n$ open, pairwise disjoint squares $Q_{1}^{n},\ldots,Q_{n}^{n}$ of side length $\left|a-b\right|/n$ plus $n-1$ vertical open line segments. Each $Q_{j}^{n}$ is trivially a translate of itself, so by our hypothesis
$$\alpha\mu(Q_{j}^{n})\geq\mu(\overline{Q}_{j}^{n}),\quad\forall j=1,\ldots,n, \ \forall n\in\mathbb{N}$$
So by subadditivity,
$$\mu\left((a,b)\times [0,\left|a-b\right|/n]\right)\leq\sum_{j=1}^{n}\mu\left(\overline{Q}_{j}^{n}\right)\leq\alpha\sum_{j=1}^{n}\mu(Q_{j}^{n})\leq\alpha\mu\left((a,b)\times(0,\left|a-b\right|/n)\right)$$
Since $\left\{(a,b)\times (0,\left|a-b\right|/n)\right\}_{n}$ is a decreasing family of open sets (of finite measure) converging to $\emptyset$, the RHS above tends to $0$ as $n\rightarrow\infty$.
