Prove $f[A] \subseteq f[B] \Rightarrow A ⊆ B$, given f is injective.

Question

I have constructed a proof for the above question, but do not know where I must use the hypothesis of injectivity:

(Note: Let $X,Y$ be sets with $A,B ⊆ X$ and $f:X \to Y $.)

Let $y∈f[A]$; then, there exists $a∈A$ s.t. $f(a) = y$. Since $f[A]⊆ f[B]$, then, we also have $y∈f[B]$ and so, $f(a)∈ f[B]$. We conclude $a∈B$. Thus, $a∈A \Rightarrow a∈B$, so $A⊆B$.

Answer 1

Suppose that $x\in A$. Then $y=f(x)\in f(A)$ and because $f(A)\subseteq f(B)$ we have that $y\in f(B)$. So there exists $w$ in $B$ such that $f(w)=y=f(x)$. But $f$ is injective, whence $w=x$ and $x\in B$ as desired.

Answer 2

Attempt:

Assume $B \subset A$ ($B$ proper subset of $A$). ($A$ \ $B \not = \emptyset$)

There exists an $a \in A$ \ $B.$

$f(a) \in f(A$ \ $B) \subset f(A)$, i.e.

$f(a) \in f(A)$;

since $f$ is injective: $f(a) \not \in f(B)$,

contradicts the assumption $f(A)\subset f(B).$