In Willard's "General Topology", exercise 9E (p67 of the Dover edition), he writes:
A decomposition $\mathscr{D}$ of a space $X$ will be called finite iff only finitely many elements of $\mathscr{D}$ have more than one point. (Typically, $\mathscr{D}$ will contain only one element with more than one point.) Prove that a finite decomposition with closed elements is upper semicontinuous. Show that the restriction that the elements of $\mathscr{D}$ be closed is necessary.
While I understand the first part of the statement (a finite decomposition with closed elements is upper semicontinuous), and can easily prove it, in my proof I use only the closedness of the elements of $\mathscr{D}$ that have more than one point.
But what about singletons in $\mathscr{D}$? Willard's statement seems to include them in closed is necessary, but is there any counter-example to show that any non-closed singleton in $\mathscr{D}$ makes the theorem fail?
Edit: I add below the definition of upper-semicontinuous, as it is a somewhat exotic subject.
An upper-semicontinuous decomposition $\mathscr{D}$ is, by definition, a decomposition of $X$ (i.e. a partition) such that, for any element $F \in \mathscr{D}$ (such element is, by definition of a decomposition, a subset of $X$, and it can be a singleton if it contains only one point), and for any open $U$ of $X$ such that $F \subset U$, there exists a saturated open $V$ of $X$ such that $F \subset V \subset U$ (saturated meaning it is an union of elements of the decomposition $\mathscr{D}$).
Edit2: here is my proof.
Take $F \in \mathscr{D}$ and take $U$ an open set such that $F \subset U$. If $U$ is saturated, then we're done. Suppose that $U$ is not saturated, we will transform it to get a saturated open set $V$. $U$ not saturated means that there is at least one element $M$ of $\mathscr{D}$ which is partly inside and partly outside $U$: $M \cap U \neq \varnothing \wedge M \cap (X - U) \neq \varnothing$. We already know that $M \neq F$ because $F$ is fully inside $U$. So we can safely remove $M$ from $U$ without changing the property $F \subset U$. We need to show that the resulting set is still open. Note that $M$ has more than two points (else if it was a singleton it would either be fully inside $U$ or outside of it). So by hypothesis it is closed. So $X - M$ is open. So, $V := (X - M) \cap U$ is open. As there are finitely many possible such $M$ elements (by hypothesis), after repeating as necessary the process the resulting $V$ is open.
Your proof is correct. Thus Willard's assertion that a finite decomposition with closed elements is upper-semicontinuous should be interpreted in the following sense:
A finite decomposition with closed non-singleton elements is upper-semicontinuous.
This is not really surprising. Consider the decomposition $\mathscr S$ of $X$ into singletons. Then each subset $U \subset X$ is saturated independent of whether any singleton is closed or not.
An alternative proof is this. Given $U \subset X$, let $\mathscr D_U$ be the set of all $M \in \mathscr D$ such that $M \cap U \ne \emptyset$ and $M \not\subset U$. Clearly $\mathscr D_U$ cannot contain singletons, thus is a finite set of closed sets and $C_U = \bigcup_{M \in \mathscr D_U} M$ is closed and $U' = U \setminus C_U$ is saturated. Thus, if $U$ is open and $F \in \mathscr D$ with $F \subset U$, then $F \notin \mathscr D_U$ and thus $F \subset U'$, where $U' \subset U $ is a saturated open set.