Order Isomorphism from Woset onto Subset

Theorem
Let $\left({S, \preceq}\right)$ be a woset.

Let $T \subseteq S$.

Let $f: S \to T$ be an order isomorphism.

Then $\forall x \in S: x \preceq f \left({x}\right)$.

Proof
Let $T = \left\{{x \in S: f \left({x}\right) \prec x}\right\}$.

We are to show that $T = \varnothing$.

So, suppose that $T \ne \varnothing$.

Then as $\left({S, \preceq}\right)$ is a woset, by definition $T$ has a minimal element: call it $x_0$.

Since $x_0 \in T$, we have $f \left({x_0}\right) \prec x_0$.

So, let $x_1 = f \left({x_0}\right)$.

$f$ is an order isomorphism, so since $x_1 \prec x_0$, it follows that $f \left({x_1}\right) \prec f \left({x_0}\right) = x_1$.

So as $f \left({x_1}\right) \prec x_1$ it follows that $x_1 \in T$.

But $x_0$ was chosen to be the minimal element of $T$.

From this contradiction, it follows that it can not be the case that $T \ne \varnothing$.

So $T = \varnothing$ and so $\forall x \in S: x \preceq f \left({x}\right)$.