The image of an extremal element is extremal

$f(x) \in M$ because $M$ is closed under $f$, so it makes sense to ask whether $f(x)$ is extremal.

It suffices to show $f(y) \le f(x)$ for the given $y \in M$ with $y < f(x)$, since that is exactly the defining condition.

Comparability at the extremal $x$ gives $y \le x$ or $f(x) \le y$.

Suppose $y < x$.

Suppose $y = x$.

The alternative $f(x) \le y$ is impossible: $y < f(x)$ gives $y \le f(x)$ with $y \ne f(x)$, so $f(x) \le y$ together with $y \le f(x)$ would force $f(x) = y$ by antisymmetry, contradicting $y \ne f(x)$. (Transitivity alone would only give $f(x) \le f(x)$, which is no contradiction; antisymmetry is what is doing the work.) Hence $y \le x$.

In the case $y < x$, extremality of $x$ gives $f(y) \le x$, and progressivity gives $x \le f(x)$, so $f(y) \le f(x)$ by transitivity.

In the case $y = x$, we get $f(y) = f(x)$, hence $f(y) \le f(x)$ by reflexivity.

The relation $y \le x$ holds exactly when $y < x$ or $y = x$, by the definition of the strict order, so the two cases are exhaustive and $f(y) \le f(x)$ in both; therefore $f(x)$ is extremal.