What do we know? We know that if $A \subseteq B$, then the cardinality of $A$ is less than that of $B$. If $A$ and $B$ are finite and the inequality is exact, then the inequality with regard to cardinality is exact as well. So, what happens if we choose $A$ to have cardinality $n$. Can we have $B$ with cardinality $n+1$ but $B$ still being a subset of the continuum? If so, how does this help us?
Is it true that if $A\subseteq B$, then the power set of $A$ is a subset of the power set of $B$? How does this give us the result?
Given a set $A$, is there a way you can think of which allows you to create a set $A^{+}$ with the cardinality of $A^{+}$ strictly greater than that of $A$? Perhaps consider the power set. Would this work? Why?
Well, if $f$ is an injection, then $A$ is in bijection with a subset of the natural numbers. Are subsets of countable sets countable?
On what specifically would you like more elaboration? Since you didn't give your thoughts on how to do it I tried to just give you hints so that you could still find the answers for yourself. Hint: If I ask you if you can do something, the answer is yes.
– JebruhoMay 02 '14 at 04:12