Claim: Let $f: S \to T$ be an injection and $A$ be a finite subset of $S$. Then, $$ |f[A]| = |A| $$ that is, there is a bijection from $f[A]$ onto $A$.
ProofWiki proves this fact using mathematical induction (MI). But does it really need MI?
Since $f$ is injective, we can simply define $g: A \to f[A]$ such that $\forall x \in A: g(x) = f(x)$. $g$ is bijective, hence, the claim follows directly.
Is there something I am missing?
I thought the set $A$ even need not be finite.
$ \newcommand{\N}{\mathbb{N}} $ The question was caused by confusion about the notation. We denote the cardinality of finite sets as $$ A \text{ is finite} \Leftrightarrow \exists n \in \N: |A| = n $$ But we also use $|\cdot|$ and $=$, $ \le$, and $\ge$ to compare the cardinality of sets, e.g., $|A| = |I_n|$ if there is a bijection between $A$ and the initial segment $I_n = \{i \in \N: i \le n\}$.
In the OP, I thought $|f[A]| = |A|$ is denoted by the latter definition. But I eventually found that the proposition actually says that for an injection $f: S \to T$, using the latter definition, $$ \forall n \in \N: \forall A \subseteq X: [|A| = |I_{n}| \Rightarrow |f[A]| = |I_{n}|] $$ or equivalently, according to the former definition, $$ \forall n \in \N: \forall A \subseteq X: [|A| = n \Rightarrow |f[A]| = n] $$ which requires MI to prove.