The union of a sequence of infinite, countable sets is countable.

Solution 1:

Look at the sequence *

$x_{11};x_{21},x_{12};x_{31},x_{22},x_{13};x_{41},x_{32},x_{23},x_{14};…$

Within each ;; add the suffixes.

1+1 =2

2+1 = 1+2 = 3

1+3 = 2+2 = 3+1 = 4

and so on.

So for any positive integer you shall get a countable (finite) number of such combination and in each case you shall get elements of $S$. If you remove duplicate items then you shall get a set $S$. This set will be bijective with the set of natural numbers as for each natural number you shall get only a finite number of elements.

I hope it is clear now. The bold sequence is constructed by taking arrows in the matrix. In the matrix the elements of the set $E_i$ are written in arrow.