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 $y\mathrm{\ne }f(x)$, so $f(x)\le y$ together with $y\le f(x)$ would force $f(x)=y$ by antisymmetry, contradicting $y\mathrm{\ne }f(x)$. (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.