It is not immediate to me that this set would be open. The standard trick to show the cantor set is closed is because it's complement is the union of all the open intervals you removed, which is open, thus the cantor set is closed. Unfortunately, in this case the complement is a union of closed intervals which need not be closed.
It might actually be easier to just take the regular cantor set and show the only thing in it is exactly the endpoints of the intervals. But in this case you tossed those out too (and 0 and 1 are removed from the beginning).
But if you really need to prove it from scratch, I suggest thinking about ternary expansions (like the decimal expansion except with only coefficients 0,1,2 ie base 3 instead of 10 so it is like ) of the cantor set. Like when you partition it up you are remove the middle third of the interval each time which corresponds to tossing everything that can be represented with a 1 in that time's digit. Know what I mean? I think you can probably show it this way as well, but it takes some serious elbow grease proving all this stuff.