Let $(X, \mathcal{T})$ be a topological space, and $\mathcal{C} \subset \mathcal{P}(X)$ be a cover of $X$, totally ordered by inclusion.
Does $\mathcal{C}$ determine $\mathcal{T}$?
That is, if $U \subset X$, and $U \cap C \in \mathcal{T}|C$ for each $C \in \mathcal{C}$, does $U \in \mathcal{T}$? Here $\mathcal{T}|C$ is the subspace topology on $C$.
No, consider $X = \mathbb N$ and the cofinite topology $\mathcal T = \{ A : |\mathbb N \setminus A| < \infty\}$. Let $\mathcal C = \{ \{1, \dots, n\} | n \in \mathbb N\}$. Clearly $\mathcal C$ is a cover and totally ordered. But I claim that any set $U$ satisfies your property, as the induced topology on each $\{1, \dots, n\}$ is the discrete topology.
If you of course assume that $\mathcal C$ is an open cover, then your statement is true, as $U = \bigcup_{C \in \mathcal C} C\cap U$ is a union of open sets, hence open.