What's with this proof that if $E_n$ is countable, then $\bigcup E_n$ is also countable? I'm a student starting Rudin's Principles right now.
I'm confused by the proof that if $E_n$ is countable, then $\bigcup E_n$ is also countable. Here it is, in image format since I don't know how to do this in TeX.

Now, as I understand the proof, since we can "index" the members of $\bigcup E_n$ like $$x_{11},x_{21},x_{12},x_{31},x_{22},x_{13},x_{41},\ldots$$ and get all the members of $\bigcup E_n$, we have a countable set. But I cannot see what's wrong with just writing it out like $$x_{11},x_{12},x_{13},\ldots,x_{1k},x_{21},x_{22},x_{23},\ldots,x_{2k},x_{31},x_{32},\ldots$$And in the event the problem is because the "$\ldots$" may not be countable (as $k\to\infty$), isn't this also the case with this diagonal line drawing thing? Eventually one of our diagonal lines will go to infinity as it has to cover $k\to\infty$ items...
Excuse me if the answer is obvious.
 A: The "table" listing the sets in the initial collection extends infinitely to right and down.  However the numbering algorithm is working along the finite diagonals (right upper to left lower).
The k-th diagonal covers k elements and the algorithm proceeds along x1,k, x2,(k-1), ... , to x(k-1),2 x1,k
and then onto the $(k+1)$-th diagonal.
Thus each diagonal uses a finite number of natural numbers for labeling.  As a result each element (or "cell") in the table is reached after a finite number of steps, and is labeled with a finite and unique natural number.  This ensures that all elements in all sets in the collection are enumerated (a bijection is constructed between the union of the sets and N).
The last paragraph in the original post is displaying a bijection from the original sets but to NxN, not to N.  While listing the original collection as such is a huge initial step, the main point of the exercise (IMO) is to come up with the bijection from NxN to N and realize they have the same cardinality.
