I would like to prove the following proposition:
Proposition. Suppose $f: X \rightarrow Y$ is an injective function. Then $|B| = |f(B)|$ for arbitrary $B \subseteq X$.
Notes on notation:
We use $|B|$ to denote cardinality of some set (set $B$ in this case)
For some function $f: X \rightarrow Y$ and some $B \subseteq X$, $f(B)$ denotes the image of the set $B$, i.e $f(B) = \{f(x) \mid x \in B\}$
We will use induction.
For the base case, take $B = \emptyset$. We have $|B| = |f(B)| = 0$.
Now suppose inductively that for some $n \in \mathbb N$, it is true that for any set $B \subseteq X$ with $|B| ≤ n$, $|B| = |f(B)|$.
Take some $K \subseteq X$ such that $|K| = n$. Consider $K \cup \{x\}$ such that $x \in X$ but $x \not \in K$. Clearly, $|K \cup \{x\}| = n+1$. Now, since $x \notin K$ and $f$ is injective, $f(x) \notin f(K)$. And by inductive hypothesis we know that$|f(K)| = n$, implying $|f(K \cup \{x\})| = n +1$. $\Box$
Is it correct?
Any alternatives? (preferably using more intuitive approach)
Answer to 1. No. In your argument you are assuming that $B$ has to always be a finite arbitrary subset of $X$, yet the statement asks for a proof of arbitrary subsets $B$ of $X$ without any restriction on their cardinality.
Answer to 2. Suppose $f : X \to Y$ is injective and let $B \subseteq X$. Consider the restriction of $f$ to $B$, namely the function $f|_B:B \to f(B)$ given by $f|_B(x) = f(x)$ for all $x \in B$. By construction, $f|_B$ is surjective, and I claim that it is also injective. Indeed, if $x, y \in B$ and $x \neq y$, then $f(x)\neq f(y)$ since $f$ is injective, and hence $f|_B(x) \neq f|_B(y)$ by definition of $f|_B$, and $f|_B$ is thus injective. Therefore $f|_B$ is a bijection; in particular we have that $|B| = |f(B)|$ by definition of two sets having the same cardinality, so we're done.