To prove that if $\sigma$ is true in all algebraically closed fields of large enough characteristic $p$, it is true in all algebraically closed fields of characteristic $0$, we can proceed as follows.
$1$) All algebraically closed fields of characteristic $0$ are elementarily equivalent. This can be proved by using the fact that any two algebraically closed fields of characteristic $0$ and cardinality $\kappa\gt \omega$ are isomorphic.
$2$) So if $\lnot\sigma$ is true in some algebraically closed field of characteristic $0$, then $\lnot\sigma$ is true in all algebraically closed fields of characteristic $0$. Now we use a compactness argument. There are various ways to phrase it.
To the theory $T$ of algebraically closed fields, add the set
$$A= \{\lnot(1+1=0), \lnot(1+1+1=0),\dots\}$$
of axioms. If $\lnot\sigma$ is a theorem of the theory $T'$ with axiom set $T\cup A$, then it is a theorem in some finite subtheory $F$ of $T'$. But $F$ mentions only finitely many of the special axioms in $A$, so $\lnot\sigma$ is true in all algebraically closed fields of large enough finite characteristic.
The converse is simpler. If $\sigma$ holds in all algebraically closed fields of characteristic $0$, then $\sigma$ is a theorem of the theory $T'$, so is a consequence of a finite collection of axioms of $T'$. These axioms are satisfied by any algebraically closed field of large enough characteristic.