Cantor–Bernstein–Schröder theorem

Let $f : A \to B$ and $g : B \to A$ be the two given injections. The idea is to trace, for each element, the chain of ancestors obtained by repeatedly undoing $f$ and $g$, and to build the final bijection piece by piece depending on where each chain "starts".

For $a \in A$, define its backward chain $a, g^{-1}(a), f^{-1}(g^{-1}(a)), \ldots$, continuing as long as the needed inverse is defined, and symmetrically for $b \in B$. Every element's chain either goes back forever, or stops at an element of $A$ with no $g$-preimage, or stops at an element of $B$ with no $f$-preimage. This partitions $A$ into three parts $A_A$ (chain stops in $A$), $A_B$ (chain stops in $B$), $A_\infty$ (chain never stops), and likewise partitions $B$ into $B_A, B_B, B_\infty$.

On chains that stop in $A$ or never stop, $f$ itself already gives a bijection from that part of $A$ onto the corresponding part of $B$ (since these elements were reached by injections and nothing "runs out" on the $B$ side first). On chains that stop in $B$, it is $g$ that gives a bijection from the corresponding part of $B$ back onto that part of $A$, so its inverse $g^{-1}$ gives a bijection from that part of $A$ onto that part of $B$.

Define $h : A \to B$ by $h(a) = f(a)$ if $a \in A_A \cup A_\infty$, and $h(a) = g^{-1}(a)$ if $a \in A_B$. Since the three pieces are disjoint and each piece maps bijectively onto its matching piece of $B$ ($f$ onto $B_A \cup B_\infty$, $g^{-1}$ onto $B_B$), the combined map $h$ is a bijection from all of $A$ onto all of $B$, proving $|A|=|B|$.