According to my understanding, if you can find a counting scheme where eventually any element of a set will be counted, you prove the countability of a set in some loose sense. Examples: proving coutability of cross product of two countable sets or countable union of countable sets. In case of reals in the interval $[0,1)$, I was thinking that we can too find a counting scheme. First let us count the decimals having only one free place (starting from the decimal to the right) to have any digit and rest all zero like 0.0,0.1,0.3 etc. Second let us have two free places, so the possibilities will be 0.01,0.89 etc. Similarly 3 places and so on. Also let us take care to remove any numbers having insignificant zeros as they are redundant, like 0.5600 etc..In each steps we have finite no of values to count and if we follow this scheme, eventually all the reals will be included(or will they be?).If not! which part of the proof changed from the countable union of countable set proof where everything will be included eventually?
And what about this other scheme : "First count the decimals that have only one nonzero digit, then those with at most two nonzero digits, and so on; we have then broken the set into a countable collection of countable sets"?
I think I have found an answer. There will be many reals which have infinite no. of non zero numbers in decimal places which will never by exhausted by my counting schemes making the proofs invalid.