Is there anything to prove in this corollary? Show that if  $B$ is not finite and $B\subset A$, then A is not finite.
I mean the statement is very trivial, but I'm having an issue actually writing what I would deem a good proof of this. The only idea I have is of letting $x\in B$ and  then show it is in $A$, but even that is trivial because I'm given that in my assumption. 
 A: As $B$ is not finite, there is an injective function $f:\mathbb{N}\to B$ (by definition).  Define $g:\mathbb{N}\to A:x\mapsto f(x)$.  Now $$g(x)=g(y)\iff f(x)=f(y)\iff x=y$$ by injectivity of $f$.  Hence $g$ is an injection to $A$.  Hence $A$ is not finite by definition.
A: Suppose there's an injective map $A\stackrel{\phi}{\hookrightarrow} n\in\mathbb{N}$
Since $B$ is infinite, there's an injective map $n\stackrel{\theta}{\hookrightarrow} B$ (I think this is the only point where one has to be careful not to use the axiom of choice)
But the inclusion $B\subseteq A$ gives us an injective map $B\stackrel{id_B}{\hookrightarrow}A$
But then we have a chain of injective maps
$$n\stackrel{\theta}{\hookrightarrow}B\stackrel{id_B}{\hookrightarrow}A\stackrel{\phi}{\hookrightarrow}n$$
But then, by Cantor-Bernstein's theorem, there exists a bijection $B\stackrel{\gamma}{\longleftrightarrow}n$. Absurd.
A: Think of the inclusion map, $\iota: B \to A, \iota(b) \mapsto b$. This is definitely an injection. Therefore $\operatorname{card} B \leq \operatorname{card} A$.
