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