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Let consider a Galois transcendental field extension $T/K$, therefore for each subextension $L$ of $T/K$ we have $T^{\operatorname{Aut}(T/L)} = L$.

My question is how to prove that this conditions already imply that $T$ is a field of characteristic $0$.

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Assume $char(K)=p>0$ and let $x\in T$ be transcendental over $K$ and consider $L=K(x^p)$. If $\alpha\in Aut(T)$ fixes $L$, i.e. $\alpha(x^p)=x^p$, then it also necessarily fixes $x$ because $x$ is the unique $p$-th root of $x^p$. Thus $T^{Aut(T|L)}$ is strictly bigger than $L$.

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