Looking for a condition on a subspace of a Hausdorff topological space

Question

Let $X$ be a Hausdorff topological space and $C$ be a closed subspace of $X$ with induced topology. I am looking for some topological conditions on the space $C$ such that if $U_1$ and $U_2$ are two open (or closed) subsets of $X$ with $C\cap U_1=C\cap U_2$, then $U_1=U_2$. I tried to show that if $C$ is a dense subspace of $X$ then the above result is true, but I could not to do. My professor said that there are at least three such properties.

Answer 1

The proof goes without saying if $C$ is dense in $(X,\tau)$, for by definition $\overline{C}=X$ and since $C$ is closed $C=\overline{C}$ or rather $C=X$. Does this answer your question?

Answer 2

Suppose $C$ is a closed set of $X$ with the property that for all open sets $U_1 , U_2$ of $X$ we have $$U_1 \cap C = U_2 \cap C \rightarrow U_1 = U_2$$ then $C = X$.

Suppose $C \neq X$, then define $U_1 = X\setminus C \neq \emptyset$ which is open as $C$ is closed. Also define $U_2 = \emptyset$. Then $U_1 \cap C= \emptyset = U_2 \cap C$. But $U_1 \neq U_2$, contradiction. So $C= X$.

As formulated now, there aren't three topological conditions on $C$ that will ensure the property. Just being $X$.