EDIT: I incorrectly interpreted the original question. The bounty has been awarded based on the (incorrect) interpretation I wrote here. The final version is posted here.
Consider the function $u : \mathbb{R}^2 \to \mathbb{R}$ that the following properties are satisfied:
Every level curve is a function from $\mathbb{R}$ to $\mathbb{R}$. (That is, a level curve $U(x,y)= c$ can be written as $y = f(x)$ for some $f : \mathbb{R} \to \mathbb{R}$.)
$\forall$ $p \in \mathbb{R}^2$, the sets $L(p)$ and $U(p)$ are closed. (Definition below.)
Definition: For $p \in \mathbb{R}^2$, define the lower contour of $p$ as $L(p) := \{(x,y) \in \mathbb{R}^2 : u(x,y) \leq u(p)\}$ and the upper contour of $p$ as $U(p) := \{(x,y) \in \mathbb{R}^2 : u(x,y) \geq u(p)\}$.
Question: Let $X = \mathbb{R}^2$. Given $u(\cdot, \cdot)$ such that one of the level curves $y = f(x)$ is discontinuous at $x = t$, construct (or show the existence of) two points $p, l \in X$ such that $l$ is a limit point of both $L(p)$ and $U(p)$ but is contained in exactly one of the two (contour sets). (The other level curves may or may not be discontinuous.)
I posted a question on Economics SE that I could solve only partially. This is essentially the same question presented in a different manner. Here's the original question:
Let $X = \mathbb{R}^2$. Suppose $\succeq$ denotes a continuous preference relation. If every indifference curve can be represented by functions from $\mathbb{R}$ to $\mathbb{R}$, will it mean the ICs will be continuous functions?
If you want to see my attempt (and/or post an answer), please click on this.
[Edit : I claimed something false, the following is wrong.]
The fact that $L(p)$ and $U(p)$ are closed implies that the function $f$ associated with $c = u(p)$ is continuous. Indeed, if $f$ is such that $\{ (x, y) \in \mathbb{R}^2, u(x, y) = u(p) \} = \{ (x, f(x)), x \in \mathbb{R} \}$, you can write : [the next equality is false]
$$ L(p) = \{ (x, y) \in \mathbb{R}^2, y \leq f(x) \} $$
and $L(p)$ is closed if and only if $f$ is upper-semicontinuous. (This is a well-known property of upper-semicontinuous functions : see for instance here.)
Now, the same reasoning with $U(p)$ implies that $f$ is also lower-semicontinuous. Therefore, $f$ is actually continuous. (I think this proves the claim you quote ?)
EDIT : Second attempt : If $l \in X$ is a limit point of both $L(p)$ and $U(p)$, since both sets are closed, you have $l \in L(p)$ and $l \in U(p)$. Therefore (and this is actually independant of the continuity of level functions $f$) you cannot find an $l$ as in your question.