I've been reading the simple parts of Thomas Jech's book on the Axiom of Choice and came across the following proof on page 141. The proof assumes a mathematical model without the axiom of choice.
Lemma. $X$ is D-finite if and only if $X$ does not have a countable subset.
Theorem. There is a set $S$ of real numbers and a real number $a \not \in S$ such that a is in the closure of $S$ but there is no sequence $(x_n)^{\infty}_{n=0} \subset S$ such that $(x_n)^{\infty}_{n=0} \longrightarrow a$.
Proof. Let $D$ be a D-finite infinite set of reals. We claim that the set $D$ must have an accumulation point. To prove this claim suppose that all $d \in D$ are not accumulation points, then let $\{I_n:n=0,1,...\}$ be a fixed enumeration of open intervals with rational endpoints. Assign the least $n$ such that $I_n \cap D={d}$ to each $d \in D$. This makes $D$ countable and by the above lemma gives us our contradiction. So $D$ has an accumulation point $a$. Let $S=D \setminus\{a\}$ and since $S$ is D-finite, every convergent sequence in $S$ is eventually constant.
Could somebody explain this proof? Why is the last sentence true/ Why does this mean every convergent sequence is eventually constant?
I think the following theorem is important
Theorem. There is a model of ZF which has an infinite set of real numbers without a countable subset