WE have three responses for you.
I would say that a real number that has all ones in it has really only two components to it: the number of digits in front of the decimal and the number behind the decimal. If you consider the number of digits in front of the decimal as the integer x and the number behind the decimal as the number y, then the set can be mapped one-to-one to the first quadrant in the x,y plane of integers.
Now if that set is countable, then your original set must be countable. I'll leave it to you to explore this.
Stephen La Rocque.>
I am not positive about which numbers you are looking at (i,e, I think you are including finite decimals, with the 0's at the end left off) and suspect the number can have 1's on both sides of the decimal, but anticipate there are no 0's scattered among the 1's.
in any case, the key is the try to imagine putting the numbers into a list. Perhaps a few lists, going down in the decimal places, and up in the other direction and ... .
There are ways to turn a countable list of countable lists into a single large countable list, so you can work these out. Finding a way to list them all will be the desired option IF they are countable.
IF you really cannot find a way to list them, then you will need to consider a proof (by contradiction) that there is no list (what is usually done for the set of all real numbers).
Why not list them all? First list those consisting of one 1. (There will be two of them, 1 and .1.) Then those with two 1's (three of them), then with three 1's, and keep going. The important question to ask yourself: can you count them or not?