A scheme $X$ is quasi separated if intersection of two quasi compact subsets is a quasi compact subset.
Question is to prove that :
A scheme $X$ is quasi separated if intersection of two affine open subsets is a finite union of affine open subsets.
Let $U,V$ be quasi compact subsets of $X$ and let $\{\text{Spec}(A_i)\}$ be an open cover for $U\cap V$.
We have $\text{Spec}(A_i)\subseteq U$ for every $i$. So, $\bigcup \text{Spec}(A_i)\subseteq U$. We complete this cover by adding some more open subsets namely $\{\text{Spec}(B_j)\}$. We thus have $$U=\left(\bigcup \text{Spec}(A_i)\right)\bigcup \left(\bigcup \text{Spec}(B_j)\right).$$ Similarly for $V$ we will add some more open subsets, namely $\{\text{Spec}(C_k)\}$ and get an open cover for $V$. We thus have $$V=\left(\bigcup \text{Spec}(A_i)\right)\bigcup \left(\bigcup \text{Spec}(C_k)\right).$$ As $U$ is quasi compact, this open cover has a finite subcover. Suppose we have $$U=\text{Spec}(A_1)\bigcup \text{Spec}(B_1).$$ As $V$ is quasi compact, this open cover has a finite subcover. Suppose we have $$U=\text{Spec}(A_2)\bigcup \text{Spec}(C_2).$$ We thus have, $$U\cap V=(\text{Spec}(A_1)\cap \text{Spec}(A_2))\bigcup(\text{Spec}(A_1)\cap \text{Spec}(C_2))\bigcup(\text{Spec}(B_1)\cap \text{Spec}(A_2))\bigcup(\text{Spec}(B_2)\cap \text{Spec}(C_2)).$$ clearly, first three terms in the union is contained in $$\text{Spec}(A_1)\bigcup \text{Spec}(A_2).$$ It is the last term which we have to thik about. As it is given that intersection of two affine open subsets is finite union of affine open subsets, we have $$\text{Spec}(B_2)\cap \text{Spec}(C_2)=\text{Spec}(D_1)\bigcup\text{Spec}(D_2).$$ Clearly, $\text{Spec}(B_2)\cap \text{Spec}(C_2)\subseteq U\cap V$ i.e., $\text{Spec}(D_1)\bigcup \text{Spec}(D_2)\subseteq U\cap V=\bigcup \text{Spec}(A_i)$.
As $\text{Spec}(D_1)$ is quasi compact and $\text{Spec}(D_1)\subseteq \bigcup \text{Spec}(A_i)$ we can assume that $$\text{Spec}(D_1)=\text{Spec}(A_3)\bigcup \text{Spec}(A_4).$$ As $\text{Spec}(D_2)$ is quasi compact and $\text{Spec}(D_2)\subseteq \bigcup \text{Spec}(A_i)$ we can assume that $$\text{Spec}(D_2)=\text{Spec}(A_5)\bigcup \text{Spec}(A_6).$$ So, $$\text{Spec}(D_1)\bigcup \text{Spec}(D_2)\subseteq \text{Spec}(A_3)\bigcup \text{Spec}(A_4)\bigcup \text{Spec}(A_5)\bigcup \text{Spec}(A_6).$$ Thus, $$U\cap V\subseteq \text{Spec}(A_1)\bigcup \text{Spec}(A_2)\text{Spec}(A_3)\bigcup \text{Spec}(A_4)\bigcup \text{Spec}(A_5)\bigcup \text{Spec}(A_6).$$ So, given an open cover $\text{Spec}(A_i)$ for $U\cap V$ we have a finite sub cover $\{\text{Spec}(A_i)\}_{i=1}^6$ for $U\cap V$.
Thus, $U\cap V$ is quasi compact.
I would like to know if this justification is correct.