Isolated points are open if $|x| \geq 2$? 
Theorem: Define $X$ to be a topological space with $|X| \geq 2. $ Then $x \in X$ is an isolated point$\iff$ $\{ x \}$ is open.

I am reading this and the proof proceeds with a neighbourhood $U$ of $x$ and showing that $U = \{ x \}.$ which uses the fact that $|X| \geq 2.$ From $U \cap X - \{ x \} = \emptyset \implies X - \{ x \} \neq \emptyset$
But I don't understand why can't we say that $U$ is open, and $X \in \tau$, so that is open, then $\{ x \} = U \cap X$, the intersection of two open sets, so it is open. What is wrong with this argument?
Def: Define $X$ to be a topological space. A point $x \in A \subset X$ is an isolated point of $A \iff \exists$ an open neighbourhood of $U$ of $x$ such that $U \cap A = \{ x \}$
 A: I'm going to restate the arguments in my own words for the sake of clarity for myself (and possibly others)
By the definition of an isolated point of a set $A$ we have that there must exist some open neighborhood of $x$, call it $U$, such that
$$
U \cap A = \{ x \}
$$
Now we take $A = X$ so then we get that
$$
U \cap X = \{x\}
$$
and since $U$ and $X$ must be open we have that $\{x\}$ is open.
So, yes, I agree with your argument.

As for how the book is presenting theirs by saying let $U$ be the open neighborhood of $x$ such that
$$
U \cap X = \{x\}
$$
(of course this exists since $x$ is an isolated point) then we have
$$
U \cap X \setminus \{x\} = \emptyset
$$
but since $\lvert X \rvert \ge 2$ we know $X \setminus \{x\} \neq \emptyset$ so thus we must have $U = \{x\}$, thus $\{x\}$ is open.

I believe the reason they present this argument this way is because you can fairly easily see both sides of the $\iff$ - i.e. you can start from the end and see how to go backwards fairly easily.
I think they only prove this for $\lvert X \rvert \ge 2$ since for $\lvert X \rvert = 1$ the statement is almost trivially true.
