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Consider $X= Spec( \mathcal O_K)$ and an open subset $ U \subset X \quad (U\neq \emptyset, X)$.
Take two copies $U'\subset X',U''\subset X''$ of the above and glue them along the identity $U'\to U''$.
You will obtain a scheme $\bar X$ that is covered by the two different open subschemes $X',X''$ each isomorphic to $\mathcal O_K$.
The scheme $\bar X$ is integral, normal (since the open subschemes $X',X''$ which cover it are), it strictly contains two copies of $\mathcal O_K$ and of course is not affine since it is not separated.

Edit When I wrote this answer I had started to suspect that no separated example existed , but couldn't prove it. Fortunately Qing has now provided a (beautiful) proof and I'm very grateful to him (as is everybody else, I guess).

Consider $X= Spec( \mathcal O_K)$ and an open subset $ U \subset X \quad (U\neq \emptyset, X)$.
Take two copies $U'\subset X',U''\subset X''$ of the above and glue them along the identity $U'\to U''$.
You will obtain a scheme $\bar X$ that is covered by the two different open subschemes $X',X''$ each isomorphic to $\mathcal O_K$.
The scheme $\bar X$ is integral, normal (since the open subschemes $X',X''$ which cover it are), it strictly contains two copies of $\mathcal O_K$ and of course is not affine since it is not separated.

Edit I wasn't too happy with this non-separated example when I posted it, but Qing now has proved that it is impossible to find a separated one.

Consider $X= Spec( \mathcal O_K)$ and an open subset $ U \subset X \quad (U\neq \emptyset, X)$.
Take two copies $U'\subset X',U''\subset X''$ of the above and glue them along the identity $U'\to U''$.
You will obtain a scheme $\bar X$ that is covered by the two different open subschemes $X',X''$ each isomorphic to $\mathcal O_K$.
The scheme $\bar X$ is integral, normal (since the open subschemes $X',X''$ which cover it are), it strictly contains two copies of $\mathcal O_K$ and of course is not affine since it is not separated.

Edit When I wrote this answer I suspected that no separated example existed , but couldn't prove it. Fortunately Qing has now provided such a proof and I'm very grateful to him.

Consider $X= Spec( \mathcal O_K)$ and an open subset $ U \subset X \quad (U\neq \emptyset, X)$.
Take two copies $U'\subset X',U''\subset X''$ of the above and glue them along the identity $U'\to U''$.
You will obtain a scheme $\bar X$ that is covered by the two different open subschemes $X',X''$ each isomorphic to $\mathcal O_K$.
The scheme $\bar X$ is integral, normal (since the open subschemes $X',X''$ which cover it are), it strictly contains two copies of $\mathcal O_K$ and of course is not affine since it is not separated.

Edit When I wrote this answer I had started to suspect that no separated example existed , but couldn't prove it. Fortunately Qing has now provided a (beautiful) proof and I'm very grateful to him (as is everybody else, I guess).

I say yes.

Consider $X= Spec( \mathcal O_K)$ and an open subset $ U \subset X \quad (U\neq \emptyset, X)$.
Take two copies $U'\subset X',U''\subset X''$ of the above and glue them along the identity $U'\to U''$.
You will obtain a scheme $\bar X$ that is covered by the two different open subschemes $X',X''$ each isomorphic to $\mathcal O_K$.
The scheme $\bar X$ is normal since the open subschemes $X',X''$ which cover it are  , but $\bar X$ is not affine since it is not separated.

Consider $X= Spec( \mathcal O_K)$ and an open subset $ U \subset X \quad (U\neq \emptyset, X)$.
Take two copies $U'\subset X',U''\subset X''$ of the above and glue them along the identity $U'\to U''$.
You will obtain a scheme $\bar X$ that is covered by the two different open subschemes $X',X''$ each isomorphic to $\mathcal O_K$.
The scheme $\bar X$ is integral, normal (since the open subschemes $X',X''$ which cover it are), it strictly contains two copies of $\mathcal O_K$ and of course is not affine since it is not separated.

Edit When I wrote this answer I suspected that no separated example existed , but couldn't prove it. Fortunately Qing has now provided such a proof and I'm very grateful to him.