I proved the following claim (source image), can you tell me if my proof is correct? Thanks:
Claim 4: Let $R$ be a binary relation on a set $X$, and suppose $\langle Y, \prec \rangle$ is a strict w.o. If there exists a function $rk : X \to Y$ such that $ \forall x, y \in X (x \neq y \land \langle x, y \rangle \in R \to rk(x) \prec rk(y)), \tag{*}$ then the relation $R$ is wellfounded.
where we use the following definition of wellfounded:
$\newcommand{\pair}[2]{\langle#1,#2\rangle}$
Definition 3: Let $R$ be a binary relation on a set $X$. We say that $R$ is wellfounded, if for every nonempty subset $Y \subseteq X$ there exists a $z \in Y$ such that $\pair yz \notin R$ for all $y \in Y\setminus \{z\}$. A relation $R$ is strictly wellfounded if it is wellfounded and irreflexive.
Proof:
Assume $R$ is not well-founded. Then there exists an infinite $R$-descending sequence $(x_n)$ such that $(x_{n+1}, x_n) \in R$, $n \in \mathbb N$, so that $S = \{x_n \mid n \in \mathbb N\}$ does not have an $R$-minimal element. Then $f(S)$ is a subset of $Y$ containing an infinite $R$-descending sequence $(f(x_{n+1}), f(x_n)) \in \prec$ which is a contradiction to $(Y,\prec)$ being well-founded.