Union of Sets Dense in $[0,1]$ I'm working on a past exam to help me study for my finals and I came across the following question:

Let $\varepsilon>0$ and define $A=\bigcup_{j=1}^\infty(x_j-\varepsilon,x_j+\varepsilon)$ where $x_j\in\mathbb{R}$. Suppose that $A\cap[0,1]$ is dense in $[0,1]$. Then $|A\cap[0,1]|=1$ (Lebesgue measure).

A hint is provided, suggesting the use of Lebesgue's Differentiation Theorem. On one hand, I don't see how to use the theorem, but, is it necessary? My line of thinking is as follows:
Since we're in $\mathbb{R}$ and $A$ is open (as a union of open sets), that means $A$ can be written as a countable union of disjoint open intervals, say $\bigcup_{j=1}^\infty(a_j,b_j)$. Assume the measure of $A\cap[0,1]$ is strictly less than $1$, say $1-\delta$ for some $\delta>0$. By the minimum fixed length of $b_j-a_j$, this implies there is some nondegenerate interval in $[0,1]$, which in turn implies there are elements with open sets disjoint from $A$, which is a contradiction since $A$ is dense in $[0,1]$. 
In fact, if I'm thinking about the construction correctly, it would seem to contain all but finitely many (based on the idea that $b_j-a_j\geq2\varepsilon$ for each $j$). I realize my proof doesn't generalize to any higher dimensions (although something similar might work) since an open set can't be written as a countable union of disjoint open sets in higher dimensions. Will anyone critique my proof and possibly suggest a line of proof using Lebesgue's Differentiation Theorem? Thanks!
 A: Let $\varepsilon>0$ and $\{x_j\}_{j=1}^\infty\subseteq\mathbb{R}$. Define $$A=\bigcup_{j=1}^\infty(x_j-\varepsilon,x_j+\varepsilon).$$ Then, since $A$ is an open set in $\mathbb{R}$, we can rewrite $A$ as a disjoint union of open sets, say $$A=\bigcup_{j=1}^\infty\:(a_j,b_j).$$ Observe that $b_j-a_j\geq2\varepsilon$. Now, we can order these intervals from left to right on $[0,1]$ (since $0$ and $\frac{\varepsilon}{2}$ can be in at most two different intervals, we have a "first" interval). After ordering the intervals, note that if $a_{j+1}-b_j>0$, then $A\cap[0,1]$ is not dense in $[0,1]$, a contradiction. Thus, $a_{j+1}-b_j=0$ for all $j$ where $(a_j,b_j)\cap[0,1]\neq\varnothing$. $^{(*)}$ Observe that since $2\varepsilon>0$, there exists a positive integer $N$ such that $N(2\varepsilon)>1$, thus if we take $N$ distinct points in $[0,1]$, there can be at most $N-1$ points not in $A$ (otherwise $m(A\cap [0,1])\geq N(2\varepsilon)>1$, a contradiction). Hence, $A$ contains all but finitely many points of $[0,1]$; therefore, clearly $m(A\cap[0,1])=1$, as was to be shown.
$^{(*)}$Here we could also argue that that there are at most countably many points not in the union and deduce the result from that, but we show something stronger.
