The more general fact is true. If $(C_\alpha)_{\alpha\in I}$ is a collection of closed subsets with finite intersection property, then all these subsets have a common point.
Assume $\cap_{\alpha\in I}C_\alpha=\varnothing$, then for open subsets of $C$ which we denote $U_\alpha=C\setminus C_\alpha$ we have $\cup_{\alpha\in I}U_\alpha=C$. Since $C$ is compact, we have finite collection $\{\alpha_1,\ldots,\alpha_n\}\subset I$ such that $C=U_{\alpha_1}\cup\ldots\cup U_{\alpha_n}$. Taking complements we get that $C_{\alpha_1}\cap\ldots C_{\alpha_n}=\varnothing$. Contradiction, so $\cap_{\alpha\in I}C_\alpha\neq\varnothing$
I'll give a more detailed version.
Suppose that $C_0 \supseteq C_1 \supseteq C_2 \supseteq \ldots C_{k} \ldots \supseteq C_{k+1} \ldots$, where all $C_k$ are compact non-empty (and thus closed, as we are in the reals).
Suppose for a contradiction that $\bigcap_n C_n = \emptyset$. The idea is to use that $C_0$ is compact, so we define an open cover of $C_0$ by setting $U_k = C_0 \setminus C_k$ for $k \ge 1$. Note that these are open in $C_0$ as $C_0 \setminus C_k = C_0 \cap (X \setminus C_k)$ is a relatively open subset of $C_0$ (using that all $C_k$ are closed so have open complement).
Also $U_1 \subseteq U_2 \subseteq U_3 \ldots U_k \subseteq U_{k+1} \ldots$, as the $C_k$ are decreasing.
Take $x \in C_0$. Then there is some $C_k$ such that $x \notin C_k$ (or else $x \in \bigcap_n C_n = \emptyset$), and so this $x \in U_k$ for that $k$.
This shows that the $U_n$ form an open cover of $C_0$, so finitely many $U_k$, say $U_{k_1}, U_{k_2},\ldots, U_{k_m}, k_1 < k_2 \ldots k_m$ cover $C_0$, so using the increasingness, we see hat $C_0 \subseteq U_{k_m}$. But take any $p \in C_{k_m}$ (by non-emptiness), then $p \in C_0$ and $p \notin U_{k_m}$, contradiction. So $\bigcap_n C_n \neq \emptyset$.
Best Answer
It is true that for finite $N \in \mathbb{N}$, $\cap_{k=1}^{N} C_{k} = C_{N}$. But consider $C_{n} = \left[ -\frac{1}{n}, \frac{1}{n} \right] \subset \mathbb{R}$. Taking an infinite intersection, we have that $\cap_{k \in \mathbb{N}} C_{k} = \{ 0 \}$. Here, the notion of largest index does not quite make sense. In a general topological space, $\cap_{k} C_{k}$ is non-empty because one can construct a sequence which converges in the intersection by the closed property of the intersection.