I'm reading the section on continua in Willard's General Topology. His Lemma 28.7 reads
If $K$ is a continuum and $(p,U,V)$ is a cutting of $K$, then each of $U$ and $V$ contains a noncut point of $K$.
See the end of the question for related definitions and lemmas, 28.2 in particular. What confuses me is his proof:
Suppose each point $x$ in $U$ is a cut point, introducing a cutting $(x,U_x,V_x)$ of $K$... say $U_x$ is contained in $U$... Since $\{U_x\cup\{x\}\mid x\in U\}$ is directed by inclusion, $\bigcap_{x\in U}[U_x\cup\{x\}]$ is a nonempty contiuum contained in $U$, by 28.2.
I don't believe the boldfaced claim to be true. For example, let $K$ is a Y-shaped continuum. Choose three cut points $p,x,y$ from three branches respectively, such that $x,y\in U_p$ and $p\in V_x\cap V_y$. Then $(U_x\cup\{x\})\cap(U_y\cup\{y\})$ is the intersection of the tails of two branches, so it is empty.
I managed to give a proof by choosing the cut points $x\in U$ more carefully. My question is: How to prove this lemma without such a heavy use of the axiom of choice?
My proof (a revised version of Willard's):
Fix $y\in V$, and put $(x_0,U_0,V_0)=(p,U,V)$. Suppose that a sequence $((x_\alpha,U_\alpha,V_\alpha))_{\alpha<\beta}$ of cuttings of $K$ is given for some ordinal $\beta>0$, such that $U_\gamma\cup\{x_\gamma\}\subseteq U_\alpha$ and $y\in V_\gamma$ whenever $\alpha<\gamma<\beta$. Then the intersection $A=\bigcap_{\alpha<\beta}(U_\alpha\cup\{x_\alpha\})$ is a nonempty continuum. Choose a cut point $x_\beta\in A$, provided that there is any. Denote the corresponding cutting by $(x_\beta,U_\beta,V_\beta)$, where $y\in V_\beta$. Given any $\alpha<\beta$, since $V_\alpha\cup\{x_\alpha\}$ is a continuum that does not contain $x_\beta$, $V_\alpha\cup\{x_\alpha\}$ does not intersect both $U_\beta$ and $V_\beta$. It intersects $V_\beta$ at $y$, so it is contained in $V_\beta$. Hence $U_\beta\cup\{x_\beta\}\subseteq U_\alpha$.
This recursive process must stop at some ordinal $\theta$, where $\bigcap_{\alpha<\theta}(U_\alpha\cup\{x_\alpha\})$ is a nonempty set of noncut points. We conclude that $U$ contains a noncut point. Similarly, so does $V$. Q.E.D.
Related definitions and lemmas:
A continuum $K$ is a connected compact Hausdorff space. We call $(p,U,V)$ a cutting of $K$, and call $p\in K$ a cut point, iff $K\setminus\{p\}$ is disconnected by its subsets $U$ and $V$ (noncut points are defined in the obvious way).
Willard's Theorem 28.2 reads:
Let $\{K_\alpha\mid\alpha\in A\}$ be a collection of continua in $X$ directed by inclusion. Then $\bigcap K_\alpha$ is a continuum.
Note that I think by "directed by inclusion" he actually means "directed by containment", i.e., $K_\alpha\supseteq K_\beta$ if $\alpha\leq\beta$.
Lemma 28.6 is also used:
If $K$ is a contiuum and $(p,U,V)$ is a cutting of $K$, then $U\cup\{p\}$ and $V\cup\{p\}$ are continua.