Prove that if $\Gamma \not \vdash \bot$, then there exists a valuation such that $[[\psi]] = 1$ for all $\psi \in \Gamma$.

Question

I tried to prove by the contrapositive of the implication, that is:

1. if there's a $\psi \in \Gamma$ such that $[[\psi]] \not = 1$ for every valuation, then $\Gamma \vdash \bot$.

2. I know that $v(\bot) = 0$ for every v. Then $\psi$ such that $[[\psi]] \not = 1$ for every valuation must be equivalent to $\bot$.

3. Then we can conclude that $\bot \in \Gamma$. Hence, $\Gamma \vdash \bot$.

But I'm not quite sure if step 3 is formally correct since I couldn't find "if $\psi \in \Gamma \rightarrow \Gamma \vdash \psi$ in my textbook even though I know it is intuitively correct