Using the Johansson's minimal logic which is a weaker form of intuitionistic logic, the author in here proves on page 83 that $A \implies B \land B \implies C$ implies $C$ holds whenever $C$ is a stable formula in the sense that $\neg \neg C \implies C$.
He defines the symbol
$ A \tilde{\lor} B \triangleq \neg A \implies \neg B \implies \bot$
and proves $A \tilde{\lor} B \implies (A \implies B) \implies (B \implies C) \implies C$.
How is it done?
Is it derived from $A \implies C \land B \implies C \land A \lor B \vdash C$?
Here's a sketch of a natural deduction proof (both for the result as you have stated it and as it is given in the source that you cite): we are given assumptions:
$$ \begin{array}{lll} (1) & \lnot A \implies \lnot B \implies \bot &\mbox{(i.e., $A\mathrel{\overline{\lor}} B$)}\\ (2) & A \implies B & \mbox{(or $A \implies C$. See comment above.)}\\ (3) & B \implies C\\ (4) & \lnot\lnot C \implies C & \mbox{(i.e., stability of $C$)} \end{array} $$
and we want to derive $C$. Assume $\lnot C$, i.e., assume $C \implies \bot $, then from $(3)$ we get $\lnot B$ and from $(2)$ we get $\lnot A$, so that $(1)$ gives us $\bot$. Discharging our assumption $\lnot C$, gives us that $(1)$, $(2)$ and $(3)$ imply $\lnot C \implies \bot$, i.e., $\lnot\lnot C$. But then using $(4)$, we get $C$, which is what we wanted.