By the well-known correspondence in infinite Galois theory, intermediate subfields $L/K$ of $\bar{K}/K$ corresponds to closed subgroups of $G=\text{Gal}(\bar{K}/K)$. In particular, generalising the finite case, if $\{L_i\}_i$ are two subfields corresponding to subgroups $\{H_i\}_i$, then the intersection $\bigcap_i L_i$ corresponds with the closure of the subgroup of $G$ generated by $H_i$. To prove this fact, if $\sigma$ is an element in the subgroup generated by the $H_i$, clearly it fixes every element of the intersection. But what about the closure? The topology on $G$ is Hausdorff, we can talk about (unicity) of limits, and an element of the closure is the limit of some converging sequence $(\sigma_j)_j$, each one fixing an arbitrary $x\in\bigcap_i L_i$. I want to pass to the limit, but at the moment I have not a topology on the field, and writing something like $$(\lim_j\sigma_j)(x)=\lim_j(\sigma_j(x))$$ makes no sense. Also, if I put a topology on the field for which the (linear) Galois maps are continuous (like the discrete), then I have to check if the limit of maps is exactly the map which computed in $x$ gives us the limit we are locking for.
Probably there is a simpler way to see this, but at the moment I cannot think at it.
In general I don't think limits are that helpful for understanding the topology in this kind of situation (in my very limited experience, at least)
Let $M = \bigcap _i L_i$ and $\Gamma_M$ its absolute Galois group. Then $\Gamma_M$ is a closed subgroup containing all of the $H_i$'s (if $L_i$ is fixed by some automorphism, then it definitely fixes the subfield $M$ of $L_i$) and in particular contains $\bar H$. So $\bar H$ is contained in the subgroup of all elements fixing $M$ and hence everything in it also fixes $M$.