Suppose not. Let $U, V$ be nonempty disjoint open sets such that $(a,b) = U \cup V$. Let $x \in U, y \in V$, and w.l.o.g. assume that $x < y$. Consider the set $S = \{z \in V | z > x\} \subseteq V$. Since $S$ is nonempty (because $y \in S$), and $x$ is a lower bound, so infimum of the set $S$ exists; call it $\alpha$. Note further that $\alpha \in (a,b)$ because $x \leq \alpha \leq y$.
Now, if $\alpha \in U$, then since $U$ is open, there is ball of some radius (say, $\epsilon$) centered at $\alpha$ in $U$, which is also contained in $(a,b)$. But that contradicts $\alpha$ being the inf, because $\alpha+\epsilon/2$ is a lower bound for $S$, and it is strictly greater than $\alpha$.
Similarly, if $\alpha \in V$, we again have a similar situation, and we can find an element in $S$ which is smaller than $\alpha$. A contradiction.
Thus we get a contradiction either way. So such a decomposition into sets is not possible. This concludes the argument.
It deviates from the standard proof, but only a little. It was graded as incorrect. So I am wondering what is wrong with this argument?
This is indeed a correct argument as confirmed via regrading. Thank you.