Integral extensions are transitive

If $A=0$, then $1_A=0_A$, so every unital image of $A$ is the zero ring; hence $B=0$ and then $C=0$, making the composite integral trivially. For the rest of the proof assume $A\mathrm{\ne }0$.

Let $c\in C$. By [L1], the element $c$ is integral over $B$, so there is a monic equation $c^n+b_{n-1}{c}^{n-1}+\cdots +b_0=0$ with $b_i\in B$. Again by [L1], each coefficient $b_i$ is integral over $A$, so [L2] makes the $A$-subalgebra $D:=A[b_0,\dots ,b_{n-1}]$ module-finite over $A$.

The same equation for $c$ has coefficients in $D$, so $c$ is integral over $D$. Because $D$ is an $A$-subalgebra of $B$, it contains the image of $1_A=1_B$, so it is nonzero. Therefore [L4] makes $D[c]$ a finitely generated $D$-module, and then [L3] gives that $D[c]$ is a finitely generated $A$-module.

The ring $D[c]$ is a faithful module over the subring $A[c]$, and step 2.1 shows that this faithful $A[c]$-module is finitely generated over $A$. By [L4], $c$ is integral over $A$. Since $c\in C$ was arbitrary, the map $A\to C$ is integral.