I'm reading Kuratowski's "Set theory", and here is a question from the Chapter 6.
Consider a linearly ordered set $A$ and its cuts, i.e., such pairs $\langle X,Y\rangle$ of $A$'s subsets that $X=Y^-$ and $Y=X^+$, where $X^+=\{a \in A:\forall (x\in X):a \geq x\}$, $Y^-=\{a \in A:\forall (y\in Y):a \leq y\}$.
Example of a cut. $\langle\{a\}^-,\{a\}^+\rangle$ (for all $a \in A$)
Let us call $A$ continuous, if for all cuts $\langle X,Y\rangle$ the condition $X \neq 0 \neq Y$ implies $X \cap Y \neq 0$. How can one prove that if $A$ is infinite and continuous, then its cardinality is $\geq \mathfrak{c}$ ?
P.S. Actually, I think the statement is wrong because the set $\mathbb{N}$ seems to be continuous according to these definitions.
It appears you are using - and + to mean the lower and upper sets.
In your example, X = {a}$^-$ and Y = {a}$^+$.
Thus X = Y$^-$ = {a}$^{+-}$ = L, the whole linear order.
So a is the top of L.
Also Y = X$^+$ = {a}$^{-+}$ = L.
So a is the bottom of L which has been reduced to a singleton.
The correct definition of a cut (X,Y) is
X $\cup$ Y = L and for all x in X, y in Y, x <= y.
Whereupon the example you gave is correct.
Let L be continuous.
If x < y and there is no point between them,
then ({x}$^-$,{y}$^+$) is a cut by disjoint sets.
Thus L is order dense.
Assume L = { $a_1, a_2,.. $} is countable.
Pick $a_1, a_2$ so $a_1$ < $a_2$.
Let $b_1$ be the first a with $a_1 < b_1 < a_2$.
Let $b_2$ be the first a with $b_1 < b_2 < a_2$.
Let $b_3$ be the first a with $b_1 < b_3 < b_2$.
Let $b_4$ be the first a with $b_3 < b_4 < b_2$.
Let $b_5$ be the first a with $b_3 < b_5 < b_4$.
Let $b_6$ be the first a with $b_5 < b_6 < b_4$.
Continue this interweaving though all the a's.
Let X = the lower set of $\cup${ b_{2n} : n in N }.
Let Y = the upper set of $\cup${ b_{2n+1} : n in N }.
Show (X,Y) is a cut and X and Y are disjoint.
Conclude L is not continuous.