Your construction might not be correct (though there may be a not obvious equivalence between your construction and the true construction, or maynot and with an obvious counter-example). First problem is that it may not well-defined that what is the large field in $A$. For example $A=k'\times k''$ for $k'\ncong k''$ non-trivial field extensions. And also there is no obvious indication that your $K$ is finite type over $k$.
I don't know whether Milne thought the construction is absolutely trivial. I must complain that it causes lots of difficulty to readers by not writing down the construction or even one reference.
The construction is based on Stack 0316.
Let $\bar{k}$ be the separable closure of $k$ and $\pi_0(X_\bar{k})$ be the set of irreducible components of $X_\bar{k}$. Then $\pi_0(X_\bar{k})$ has an $\mathrm{Gal}(\bar{k}/k)$ on it. Since we are talking about scheme of finite type, $\pi_0(X_\bar{k})$ is finite. For $X_1, X_2 \subset X$ disjoint, the orbit of $\pi_0((X_1)_\bar{k})$ and $\pi_0((X_2)_\bar{k})$ are disjoint. In fact by Lemma 038B the union of an orbit of $\mathrm{Gal}(\bar{k}/k)$ can be descended to a subscheme of $X$. Hence orbits correspond exactly to connected components of $X$.
We now look at one orbit and assume $X$ is connected. In view of Lemma 038D, we can take the kernel $N$ of $\mathrm{Gal}(\bar{k}/k) \rightarrow \mathrm{Perm}(\pi_0(X_\bar{k}))$. Then $\bar{k}^N=:k'$ satisfies that $X_{k'}$ has a trivial Galois action on it. By Lemma 038D(1), trivial Galois action is equivalent to geometrically connectedness in each components. So $k'$ is the minimal field extension that makes $X$ geometrically connected in each components and is the $\pi_0(X)$ we want to construct. In fact the action of $\mathrm{Gal}(\bar{k}/k)$ on $\pi_0(X_\bar{k})$ can be identified with the action on $\bar{k}\otimes_kk'$.
In view of Lemma 038D(2) and the fact that the action of $\mathrm{Gal}(\bar{k}/k)$ on $\pi_0(X_\bar{k})$ can be identified with the action on $\bar{k}\otimes_kk'$, the condition (2) of varieties of connected components is easy to shown. In fact let $k_1, k_2$ correspond to $\pi_0$ of $X$ and $Y$, then the Galois action on $X\times Y$ can be identified with that on $\bar{k}\otimes k_1 \otimes k_2$. So we get (3).
I find a gap in the construction. We must have a morphism from $X$ to $\pi_0(X)$. $X_\bar{k} = \coprod_{\sigma \in \pi_0(X_\bar{k})} (X_\bar{k})_\sigma$ is a decomposition into connected components. Every $(X_\bar{k})_\sigma$ is naturally a $\bar{k}$-scheme. Then we have a morphism $X_\bar{k} \rightarrow \coprod_{\sigma \in \pi_0(X_\bar{k})} (\bar{k})_\sigma$. Note that $\coprod_{\sigma \in \pi_0(X_\bar{k})} (\bar{k})_\sigma$ has a Galois action on it, consisting of the action on $\pi_0(X_\bar{k})$ and on $\bar{k}$ it self. Note that the morphism is compatible with the Galois action since we use the 'natural' $\bar{k}$ structure on $X_\bar{k}$. I explain the naturality in the following.
The natural structure can be seen locally. Assume $X = \mathrm{Spec}(A)$. Then $X_\bar{k} = \mathrm{Spec}(A\otimes\bar{k})$. Let $A\otimes\bar{k} = \prod A_i$. Then multiplication of $\bar{k}$ on $A\otimes \bar{k}$ gives the multiplication of $\bar{k}$ on each $A_i$. This gives the natural $\bar{k}$-algebra structure on each $A_i$ and each $\bar{k} \rightarrow A_i$ summing up to be $\prod (\bar{k})_i \rightarrow \prod A_i$. Let $\sigma \in \mathrm{Gal}(\bar{k}/k)$. Since $\sigma(la)= \sigma(l) \sigma(a)$ for $l \in \bar{k}$ and $\sigma(a) \in A$ and $(l)_i(a)_i = la$ for $a \in A_i= A_i \times 0 \times \cdots \times 0 \subset A$ where we write $(l)_i$ (resp. $(a)_i$) to indicate that it is considered as an element in the $i$-th component of $\prod (\bar{k})_i$ (resp. $\prod A_i$), we have that the morphism is compatible with the Galois action.
As the construction, $\prod_{\sigma \in \pi_0(X_\bar{k})} (\bar{k})_\sigma \simeq \pi_0(X)\otimes_k\bar{k}$ with the Galois action acting on the right tensor product component. Hence when you do Galois descent, you get $\pi_0(X)$, i.e. when you do Galois descent, $X_\bar{k} \rightarrow \coprod_{\sigma \in \pi_0(X_\bar{k})} (\bar{k})_\sigma$ becomes $X \rightarrow \pi_0(X)$.
You can come up with the idea by yourself once your read through that stack thread.