Basic topology of a 4 point set Let $X=\{1,2,3,4\}$ be given by the topology $\tau=\{\emptyset,X,\{1\},\{1,2\},\{1,3\},\{1,2,3\}\}$.
Show that any two disjoint closed sets have disjoint neighborhoods.
So I define my closed sets (as complements of the given open sets) as
$\mathcal{C}=\{X,\emptyset,\{2,3,4\},\{3,4\},\{2,4\},\{4\}\}$
So my question is, since $4$ is an element of every closed set except $\emptyset$, is there even anything to show because every closed set except for $\emptyset$ is contained in the neighborhood $X$ (because $X$ is the only member of $\tau$ who contains the element $4$). Is there even anything to show or?
 A: Since @Hossien Sahebjame already answered your question in the comments, let formulate the same answer in a slightly more general context, because why not.
Let $\bar{X} = X \cup \{a\}$ be a set and define the open sets of the excluded point topology on $\bar{X}$ to be all subsets, not containing $a$ and the whole set, so $\tau = \mathcal{P}(X) \cup \bar{X}$. In every such topological space every closed set contains the point $a$ and hence there are no non-trivial disjoint closed sets which implies that "any two disjoint closed sets have disjoint neighborhoods" by means of a vacuous truth (as pointed out by you and in the comments already).
Now, having a coarser topology $\sigma \subset \tau$ can only reduce the number of closed sets; therefore all topologies that are coarser than an excluded point topology have this property. Since the topology you described is coarser than the excluded point topology $\bar{X} = \{1,2,3\} \cup \{4\}$, this completes this unnecessarily overcomplicated "answer".
