$\bigcup_{i=0}^\infty I_i $ not superset of $[0,1]$? Index all rational numbers in $[0,1]$ from 1 to infinity. For each rational number $q$ in $[0,1]$, form a set $I_i=[q-\frac\epsilon{2^i},q+\frac\epsilon{2^i}]$, where $\epsilon$ is a small number, maybe $\epsilon=0.01$. Now consider the set $$A=\bigcup_{i=0}^\infty I_i $$ Set $A$ seems to be a superset of $[0,1]$ because for any number $x\in[0,1]$, there is a rational number that is very close to it. So $x\in I_i$ for some $i$. But the measure of $A$ is $\epsilon$ and the measure of $[0,1]$ is $1$. Does this mean there must be numbers in $[0,1]$ but not $A$? If so, what are those numbers or how to construct them? Also why is the argument for $x\in I_i$ for some $i$ incorrect?
 A: The argument for $x\in I_i$ is incorrect because you didn't specify "how very close" $x$ is to the rationals. There are (lots of) irrational numbers in $[0,1]$ that are not inside any $I_i$. For example, it is reasonably easy to see that you can manage to choose your indexing of the rationals in such a way that $\sqrt{2}/2$ is never inside any of the $I_i$. It just happens that this must be the case for an uncountably large number of irrationals, not only $\sqrt{2}/2$.
The way to construct these irrationals depends on your particular manner of enumerating the rationals. I said above that "it's easy to see that" you can enumerate the rationals in such a way that $\sqrt{2}/2$ is never inside any $I_i$. Well, it is indeed easy. Start with any enumeration $r_i$ of your choice. If the distance from $r_1$ to $\sqrt{2}/2$ is greater than $\epsilon/2^1$, take $q_1=r_1$. Otherwise, try the next $r_i$ until you reach an $r_i$ that is outside the interval $\left[\sqrt{2}/2-\epsilon/2^1,\sqrt{2}/2+\epsilon/2^1\right]$ (you eventually will). Now that you have set $q_1=r_i$, remove this $r_i$ from your original enumeration, and repeat the same algorithm to choose $q_2=r_j$ such that $q_2$ is outside the interval $\left[\sqrt{2}/2-\epsilon/2^2,\sqrt{2}/2+\epsilon/2^2\right]$. This way, you will end up choosing all of the $q_n$, and $\sqrt{2}/2$ will never be inside any of the $I_i$.
Also, note that the measure of $A$ is less than $\epsilon$, not equal to.
A: There are irrational numbers which are in none of the $I_i$. For convenience let me replace your $\varepsilon$ by $\delta$, and I will list the rationals in $[0,1]$ as $\{ q_i \}_{i=1}^\infty$.
You do have that for every irrational $x$ and every $\varepsilon > 0$, there is a rational $q$ with $|x-q|<\varepsilon$. What you would need is not quantified the same way. You would need that some rational $q_i$ satisfies $|x-q_i|<\delta 2^{-i}$. That is, when we pick a rational we are immediately given the radius around it.
So it could happen that when you pick $\varepsilon$, all of the $q_i$ within $\varepsilon$ of $x$ have such large $i$ that $\delta 2^{-i}<\varepsilon$. Indeed notice that only finitely many $i$ satisfy $\delta 2^{-i} < \varepsilon$. So we can keep $x$ out of the union in an iterative fashion. Specifically we can get a sequence $\varepsilon_k$ going to zero and an increasing sequence $N_k$ so that for each $k$ and each $i=N_{k-1},N_{k-1}+1,\dots,N_k-1$, $|x-q_i|>\varepsilon_k$ with $N_k$ chosen sufficiently large for each $k$.
Being any more concrete will become counterproductive, I think. Part of the problem is that concrete examples require you to concretely specify an enumeration of the rationals.
A: I'm going to assume your construction is as follows:

Enumerate the rationals as $(q_n:n\in\mathbb N)$ (you didn't specify this).  Fix some small $\varepsilon>0$, and let $I_i=[q_i-2^{-i}\varepsilon,q_i+2^{-i}\varepsilon]$.  Then form
  $$
A=\bigcup_{i\in\mathbb N} I_i
$$

You claim that $A$ contains $[0,1]$ as a subset.  The reasoning is:

For any real number $x\in[0,1]$, and for any $\delta>0$ there exists some rational $q\in[0,1]$ such that $|x-q|<\delta$.  We deduce that $x\in I_n$ for some $n$.

Unfortunately, that deduction is false.  The reason that it doesn't work is that although we can find some $q_n$ arbitrarily close to any $x\in[0,1]$, $n$ could be arbitrarily large itself, so the two endpoints $q_n-2^{-n}\varepsilon$ and $q_n+2^{-n}\varepsilon$ of $I_n$ could be even closer to $x$ than $q_n$.  I.e., $x\not\in I_n$.  
I'd give you an example of some $x$ that's not contained in any of the $I_i$, but it depends on the enumeration $(q_n)$ that you fix to start with.  Besides, you yourself have provided a perfectly good proof that such an $x$ exists - the set $A$ has measure $\varepsilon$, so can't contain the whole of $[0,1]$!
A: It is possible to "somewhat constructively" produce a number not in the union of the intervals $I_i$.  Actually, for convenience let me modify the construction slightly so that $I_i$ has length $10^{-i}$ instead of $2\epsilon/2^i$.  Start with
the interval $J_0 = [0,1] = [a_0, a_0 + 1]$.  $I_1$ has length $1/10$, so it can't intersect more than one of $[0,0.1]$ and $[0.9, 1]$.  Choose $J_1 = [a_1, a_1 + 0.1]$ to be $[0,0.1]$ if that doesn't intersect $I_1$, otherwise $[0.9, 1]$.
Again, the length of $I_2$ is $1/10$ the length of $J_1$, and we can choose 
$J_2$ to be either the leftmost $1/10$ of $J_1$ (if that doesn't intersect $I_2$) or the rightmost $1/10$ of $J_1$ if it does.  Proceed inductively: $J_n = [a_n, a_n + 10^{-n}]$ will be an interval of length $10^{-n}$ that doesn't intersect $I_1, \ldots, I_n$.
Now let $x$ be the limit of $a_n$ as $n \to \infty$: in fact $x$ will have a decimal expansion whose first $n$ digits after the decimal point are the same as those of $a_n$. 
A: The answers given have pointed out the flaw in the OP's argument. Let me say a bit, though, about the intuitive content of the mistake.
The statement "if $x$ is arbitrarily close to elements of $A$ (or more precisely, if $\forall\epsilon>0\exists a\in A(d(x, a)<\epsilon)$), then $x\in A$" is just saying "$A$ is closed."
Of course not every set is closed. For example, consider $A=\mathbb{R}-\{0\}$. Of course 0 is arbitrarily close to elements of $A$, but not in $A$.
The intuitive response to this is often, "Sure, but we can make $A$ closed by just adding one point; that won't effect the size of $A$." This is true. However, this is often then extended to,

We can always make $A$ closed just by adding "a bunch of single points," so that shouldn't change the size of $A$.

This is the part that is wrong, and often surprising for those seeing analysis for the first time! The closure of a set can be vastly larger than the original set, in essentially any sense of "largeness" whatsoever. 

A dual mistake is assuming that all open dense sets are big; in fact, you can make countably many open dense sets whose intersection has measure 0!
A: Here's a simple proof that no possible $\{I_i\}$ as specified by you will cover, based on counting integer points, using only the compactness of $[0,1]$ (once you know some measure theory, more general techniques supersede this, but I like the intuition this simple proof provides):
Definition: The length of an interval $I = [a,b]$ (with $a,b \in \mathbb{R}$ and $b > a$) is $\mathrm{len}(I) = b-a$.
Proposition: Suppose $\{I_i\}_{i\in A}$ is a collection of open intervals and $\sum_{i\in A} \mathrm{len}(I_i) < 1$. Then $\cup_i I_i \not\supset [0,1]$.
Proof: Given $\lambda \in \mathbb{R}$ with $\lambda > 0$, let $\lambda[a,b] = [\lambda a, \lambda b]$.
Let the number of integer points contained in any set $S \subset \mathbb{R}$ be denoted $n(S)$. Then we have $b-a-1 \leq n(I) \leq b-a+1$ for $I = [a,b]$.
Now, if we assume by way of contradiction that $\{I_i\}$ is an open cover of $[0,1]$ then there exists a finite subcover. So we may assume without loss of generality that the index set $A$ is finite, say $A = \{1, \ldots, m\}$.
Let $\sum_{i=1}^m \mathrm{len}(I_i) = c < 1$.
Then $n(\cup_{i= 1}^m \lambda I_i) \leq \sum_{i=1}^m (\lambda b_i - \lambda a_i + 1) = m + \lambda \sum_{i=1}^m \mathrm{len}(I_i) = m + \lambda c$.
Choose $\lambda > \frac{m}{1-c}$. Then $n(\lambda\cup_{i= 1}^m I_i) \leq m+\lambda c < \lambda = n(\lambda [0,1])$.
Thus $\lambda \cup_{i=1}^m I_i \not\supset \lambda[0,1]$.
Thus $\cup_{i=1}^m I_i \not\supset [0,1]$
$\square$
In your case, note that $\sum \mathrm{len}(I_i) = \epsilon = .001 < 1$.
