Firstly, for any locally ringed space $X$, there is an initial object in the category of affine schemes Spec $A$ equipped with a morphism $X \to $ Spec $A$, namely Spec $\Gamma(X,\mathcal O_X)$. In other words, the canonical map $X \to $ Spec $\Gamma(X,\mathcal O_X)$ is the initial object in the category of morphisms from $X$ to affine schemes. This indicates one way to reduce certain questions to the affine case.
Your question is one that can be so reduced. That is, for any locally ringed space, $X$ is disconnected if and only if Spec $\Gamma(X,\mathcal O_X)$ is disconnected. (Easy exercise.)
Now since $K$ is flat over $k$, we have that $\Gamma(X\times_k K, \mathcal O_{X\times_k K}) = \Gamma(X,\mathcal O_X)\otimes_k K.$
Thus we have reduced your question to the affine case. That is, we only have
to show that Spec $A$ connected implies that Spec $A\times_k K$ is connected.
Next, we will reduce to the Noetherian case. To this end, write $\displaystyle A = \varinjlim_i \, A_i$ as a direct limit of finite type $k$-algebras, so
that $\displaystyle A\otimes_k K = \varinjlim_i \, A_i\otimes_k K$.
Remember that to disconnect
Spec $A\otimes_k K$, you have to exhibit a non-trivial idempotent $e \in A\otimes_k K$. Such an idempotent has to come from $A_i\otimes_k K$ for some
$i$, and so if $A\otimes_k K$ is disconnected, so is $A_i \otimes_k K$
for some $i$. But then $A_i$ is disconnected (by the Noetherian case,
which I am assuming we know), and hence $A$ is disconnected. QED
A topological space is called quasi-separated if the intersection of any two quasi-compact open (not just any) subsets is quasi-compact. Check your definition.
Let $X$ be a scheme. Suppose the intersection of any two affine open subsets is a finite union of affine open subsets. Take two quasi-compact open subsets $U, V\subset X$. We want to show that $U\cap V$ is quasi-compact.
First prove the following easy lemmas:
Lemma 1 Every affine scheme is quasi-compact.
Lemma 2: If $Y$ is a topological space which is a finite union of quasi-compact spaces, then $Y$ is quasi-compact.
Lemma 3: Let $X$ be a scheme, then the affine open subsets form a base for Zariski topology.
Using the thrid lemma both $U,V$ are unions of affine opens. Using the fact that $U,V$ are quasi-compact, we find that $U,V$ are finite union of affine opens, say $U=\bigcup_{i=1}^n \mathrm{Spec}A_i$ and
$V=\bigcup_{j=1}^m \mathrm{Spec}B_j$. Then
$$
U\cap V= \bigcup_{i=1}^n \bigcup_{j=1}^m \mathrm{Spec}A_i\cap \mathrm{Spec}B_j $$
since the intersection of any two affine opens is a finite union of affine opens, then $U\cap V$ is a finite union of affine opens. Since every affine schme is quasi-compact (lemma 1), and since $U\cap V$ is a finite union of affine opens/quasi-compacts, then $U\cap V$ is quasi-compact (lemma 2).
Best Answer
Yes, that's sufficient. Being reduced is a local property; you can check it on the stalk at each point, or on any affine open covering. (This is a good exercise in sheaf theory; this answer gives a hint on how to prove it.) That your hypotheses are sufficient for irreducibility is proved in Stacks Project, Lemma 28.3.3 (tag 01OM). Since a scheme is integral iff it's reduced and irreducible, this proves what you want.