The set of $\{R_i\}$ is clearly not countable - each $i \in \mathbb{R}$ provides a different $R_i$, which gives us an explicit bijection between $\{R_i : i \in \mathbb{R}\}$ and $\mathbb{R}$. That's exactly what it means to have the same cardinality as $\mathbb{R}$, so the set $\{R_i\}$ has cardinality continuum.
The problem with your intuition seems to be that you're thinking of the indices $i$ as counting; if you were restricting $i$ to be $0,1,2$ and so on, then the set of $R_i$ would certainly be countable; but it would also no longer be the case that you have an additional real line going through each point in $\mathbb{R}$, just the non-negative integers.
EDIT: Based on the document you linked, it looks like the problem you're referring to is completely different from the statement you gave. In particular, it has nothing to do with copies of $\mathbb{R}$ placed intersecting a line. Here's what is described in the text: Fix the lexicographic ordering of $\mathbb{R}^2$, so that $(a, b) \prec (a', b')$ iff either $a < a'$ or $a = a'$ and $b < b'$. Suppose $f$ is a real-valued function representing this order, which I gather means that $f(a, b) < f(a', b')$ iff $(a, b) \prec (a', b')$. In other words, $f$ is trying to map $\mathbb{R}^2$ to $\mathbb{R}$ in a way that keeps things in the same order. Then the particular intervals of the form $[\inf f(a, \mathbb{R}), \sup f(a, \mathbb{R})]$ must be distinct, disjoint intervals in $\mathbb{R}$, each containing a different rational point.
The absolutely essential thing to understand is that the whole point of this example is to show that no such $f$ exists, for exactly the reason you outlined: if each of these uncountably many intervals contains a different rational, there can only be countably many. But there's no problem here: that just means that there is no function $f$ so that $f(a, b) < f(a', b')$ iff $(a, b) \prec (a', b')$. This is not terribly surprising; if you play around with it a bit, you'll see that it's hard to come up with an $f$ that even comes close to doing this.
EDIT 2: Based on your comments, I think it's worth going over the definition of "countable". A set $X$ is countable if there exists a function $f:X \to \mathbb{N}$ so that $f$ never takes any two different elements of $X$ to the same point. It is a known fact that $\mathbb{Q}$ is countable, and that $\mathbb{N}$ may be replaced with any infinite countable set in the definition of "countable". Thus, in Debreu's example, he demonstrates that his collection of intervals is countable by observing that each interval includes a different rational; this allows us to construct a function $f$ taking each interval to a different member of $\mathbb{Q}$, which is the function necessary to show countability. In your example, no such function is supplied, because the rational points in each $R_i$ are not different.
EDIT 3: In your example of ten sets of reals, you could indeed identify them with $1, 2, 3, \ldots, 10$. But the critical feature of that is that you had to start with exactly ten sets of reals. Likewise, if there were countably many $R_i$, you could identify each with a different rational; but you can't use that to prove that there are countably many, because you're already assuming it!
I think I have now explained, in detail, that your example cannot be fixed. If you think it can be, please give your "fix" explicitly: for example, it is patently obvious that you can't have the $R_i$ be disjoint intervals in $\mathbb{R}$, while still having every $R_i$ cover the point $i$. You could mimic Debreu's construction precisely, but doing so requires using a function $f$ representing the lexicographic order Debreu refers to; as Debreu's example points out, no such function exists.
To address (5) and (6): Debreu made no blunder, and he did not prove (nor intend to prove) that preferences must be continuous. Rather, he demonstrated that if you place no restriction on the preference order, then pathological cases like the lexicographic example can occur, in which there cannot be a representing function. He therefore concluded that some restriction must be placed in order to eliminate examples like the lexicographic case, and requiring continuity was the most straightforward way to do that.
Your (7) and (8) do not make sense. There obviously is a continuous $U$; the text you cited gives plenty of examples. There isn't one that matches the lexicographic case, but that's not a problem; that just means that the results of this theory do not apply in a situation in which preferences happen to be lexicographic. I have no idea what you mean by (8).
