Let $A$ and $B$ be sets. If $A$ is infinite and $A\subset B$, the $B$ is infinite. Proof by Contradiction: Suppose $A$ is infinite, $A\subset B$, and $B$ is infinite.
Since $A\subset B$, then $|A|\leq |B|$. 
Since A is infinite and $|A|\leq |B|$, then $B$ must be infinite. But $B$ is finite. Contradiction occurs.
Therefore $B$ is infinite.
Is this enough to verify the statement.
 A: The argument you provided is essentially a tautological restatement of the thing you are asked to prove in the "$\lvert A\rvert\le\lvert B\rvert$" formalism, so I doubt the person who gave you the assignment deems it a valid proof. For instance, you are not using explicitly the fact that "infinite set" has an actual meaning, other than satisfying some (true, but largely unspecified) formal property.
I'd rather go with: let $S\subsetneq A$ such that there is a bijective function $f:A\to S$. Then, $S\cup(B\setminus A)\subsetneq B$ and the function \begin{align}\overline f:B&\to S\cup(B\setminus A)\\ \overline f(x)&=\begin{cases}f(x)&\text{if }x\in A\\ x&\text{if }x\in B\setminus A\end{cases} \end{align}
is bijective.
A: Or, if you want contradiction, assume $f:B\to n$ is bijective for some natural number $n$.
Then $f|_A:A\to f(A)$ is bijective. And since $f(A)\subset n$, then $f(A)$ is itself a natural number.
This contradicts that $A$ is infinite since it shows a bijection from $A$ to a natural number.
