$k$ is a contradiction such that it belongs to a set of well-formed formulas. $\Sigma ; \lnot \alpha \vdash k $. Prove that $\Sigma \vdash \alpha$ where $\alpha$ is a well-formed formula.
After applying DT $\vdash$, $\Sigma\vdash (\lnot \alpha \rightarrow k) $ and we know that all truth assignments for $k$ is not satisfiable. But how can we conclude $\Sigma \vdash \alpha$?
You simply need to deduce $\alpha$ from $(\neg\alpha \to k)$. And that's an application of reductio (if something implies a contradiction you can deduce its negation) and then the classical double negation rule (from $\neg\neg\alpha$ you can infer $\alpha$.
How you join up the dots will depend on the particular proof system you are using.