Finite Sets, Equal Cardinality, Injective $\iff$ Surjective.

If the sets have cardinality $n$ and $f$ is injective, then the image of $f$ must be an $n$ element subset of $B$ and so equal to $B.$ If $f$ is surjective, then the preimage of each element of $B$ contains at least one element, and the preimages are disjoint. So the union of the preimages of the elements of $B$ has at least $n$ elements. Since $A$ has only $n$ elements, each preimage of an element of $B$ can contain only one element of $A.$ so that $f$ is injective.

For the statement "$f$ injective implies $f$ surjective", induct on the cardinality of $A$ and $B$.

You can check $n=0$.

For $n=k+1$, let $f:A\to B$ be injective. $A$ has at least one element $a$. Define $f^-:A\setminus\{a\}\to B\setminus\{f(a)\}$, gotten by forgetting about $a$ and its value $f(a)$. $f^-$ is injective as restricting injections preserves injectivity (check this if you don't know). Then $f^-$ is also surjective by induction hypothesis. Appending back $a$ and $f(a)$ preserves surjectivity so $f$ is surjective.

The other direction is exactly the same; replace injective with surjective everywhere.

The OP was attempting a proof to

$\quad$ (*) "show that a surjection implies an injection using induction"

Now user524154 worked on the other half and stated that (*) could be handled in exactly the same way. Well, there are some subtleties when using that technique.

Proposition: For all $n \ge 0$ and sets $A$ and $B$,
$\quad \quad \quad \quad \quad$ if $n = |A| = |B|$ and $f: A \to B$ is a surjection then $f$ is an injection.
Proof by induction:
For the base case $n = 0$ there is exactly one function, the empty graph mapping $A = \emptyset$ to $B = \emptyset$, and it is vacuously a bijection.
Step case: Assume the proposition is true for $n = k$. If $k = 0$ then the proposition holds for $n = k + 1 = 1$, since the (unique) function mapping one singleton set $A$ to another singleton set $B$ must be a bijection.
So we assume that $n = k + 1 \ge 2$ so that $B$ has at least two elements. Assume to get a contradiction that there is an element $b \in B$ such that

$\quad |f^{-1}(b)| \ge 2$

Select an element $a \in f^{-1}(b)$ and let $F = \bigcup f^{-1}(b) \setminus \{a\}$ so that the preimage is partitioned,

$\quad f^{-1}(b) = \{a\} \bigcup F $

Select an element $\hat b \in B$ such that $\hat b \ne b$;
note that there exists an $\hat a \in A$ not belonging to $f^{-1}(b)$ satisfying $\hat a \in f^{-1}(\hat b)$.

Consider the binary relation $\rho$ equal to

$\quad \text{Graph}(f) \setminus \bigr(f^{-1}(b) \times \{b\}\bigr) \; \bigcup \; F \times \{\hat b \}$

Checking, we verify that $\rho$ is actually a function $g$,

$\quad g: A \setminus \{a\} \to B \setminus \{b\}$

that is a surjective mapping between two sets with cardinality $k$ with at least two distinct elements in $g^{-1}(\hat b)$.

But this contradicts the inductive hypothesis. $\quad \blacksquare$