The title basically says it all. Given an arbitrary scheme $X$ with global sections $\mathcal{O}_X(X)$, is there an affine open subset $U$ of $X$ such that $\mathcal{O}_X(U) = \mathcal{O}_X(X)$?
Obviously the scheme $Y=\operatorname{Spec}\mathcal{O}_X(X)$ has the same global sections as $X$, but it's not clear to me that you can embed $Y$ in $X$. I have been unable to produce a counter example though.
Consider $X = \Bbb{P}^1_k,$ where $k$ is your favorite field. We have $\mathcal{O}_X(X) = k,$ and note that $X$ is integral, so that the restriction maps $\mathcal{O}_X(U)\to\mathcal{O}_X(V)$ are injective. Now, notice that any proper open subscheme is contained in one of the two copies of $\Bbb{A}^1_k$ inside $\Bbb{P}^1_k.$ This means that if $U$ is any affine open subscheme of $X,$ we have injections $k\to k[x]\to\mathcal{O}_X(U).$ In particular, this shows that the map $\mathcal{O}_X(X)\to\mathcal{O}_X(U)$ is never an isomorphism for an affine open subscheme $U\subseteq X.$