I think you're trying to do this the wrong way round. You are told that (X,T') is compact, and you want to show that (X,T) is compact.

So let U be an open cover of (X,T). The members of U are T-open subsets of X. But , so the members of U are T'-open. Therefore they form an open cover of (X,T'). So there is a finite subcover (and that's all that you need, to show that (X,T) is compact).