Let $X$ be a compact topological space. We have the following property: If $(F_{i})_{n \in \mathbb{N}}$ is a decreasing sequence of nonempty closed sets on $X$, then their intersection is nonempty
If this property true for a topological space then must it be compact? If yes why, and if not could you give me counter example and when it is true (for Hausdorff space or metric space or ...)
Such a space is not necessarily compact - a Hausdorff example would be the first uncountable ordinal $\omega_1$ with the order topology: This space is not compact (as the open cover given by the countable ordinals admits no finite subcover). Now suppose $(F_n)_{n\geq 0}$ is a nested decreasing sequence of nonempty closed sets. If any of the $F_n$ is bounded above by (say) $\alpha$, then the tail of the sequence is contained in the compact subspace $\alpha+1$ and therefore has a nontrivial intersection.
If on the other hand, each $F_n$ is unbounded, we can also show that their intersection is nonempty: Inductively we can find a sequence $$\beta_0<\beta_1<...$$ of countable ordinals such that $\beta_n\in F_n$ for each $n$. As a countable set of countable ordinals, the $\beta_n$ are bounded below $\omega_1$ and therefore converge to a limit $\beta<\omega_1$. Since each $F_n$ is closed and contains each $\beta_k$ for $k\geq n$, it also contains their limit $\beta$, which therefore lives in the intersection.
Edit: We can see that your intersection property implies that every countable open cover has a finite subcover: Suppose $U_0,U_1,...$ is an open cover of $X$. Consider the sets $$F_n=X\setminus \bigcup_{k=0}^nU_k.$$ The $F_n$ are then closed and nested (both clear). Their intersection however is $$\bigcap_{n\geq 0}\left(X\setminus \bigcup_{k=0}^nU_n\right)=X\setminus \bigcup_{n\geq 0}U_n=\emptyset,$$ since the $U_n$ are an open cover. Since $X$ has your intersection property, one of the $F_n$ must already be empty. But then $X=\bigcup_{k=0}^nU_k$ is a finite subcover.
In particular this shows that any Lindelöf space with your property must already be compact.