First, if B injects into C and B is uncountable, then C is uncountable.
Proof (here 'w', read 'omega', stands for the set of natural numbers):
Toward a contradiction, suppose B injects into C, and B is uncountable, but C is countable.
So C injects into w.
So, since B injects in C, we have B injects into w, so B is countable. Done.
Suppose S and T are uncountable.
Let y be in T.
S X {y} is equinmerous with S.
So S X {y} is uncountable.
S X {y} is a subset of S X T.
So S X T is uncountable.