No injection of a power set into finite sequences

Suppose $G:\mathcal{P}(X)\to \mathrm{Seq}(X)$ is injective. For an infinite well-ordered subset $(Y,<)$ of $X$, let $H:Y\to \mathrm{Seq}(Y)$ be the uniformly defined bijection. Put $D=\{y\in Y:H(y)\in \mathrm{ran}(G)\text{ and }y\mathrm{\notin }{G}^{-1}(H(y))\}$. The inverse here is used only at points in the range, where it is unique.

If $G(D)=H(y)$ for some $y\in Y$, the definition would give $y\in D$ iff $y\mathrm{\notin }D$. Hence $G(D)\mathrm{\notin }\mathrm{Seq}(Y)$. Its finite sequence has a first coordinate outside $Y$; let $a(Y,<)$ be that value. This rule is unique and definable from $G$ and the given order. The sequence cannot be empty, since the empty sequence belongs to $\mathrm{Seq}(Y)$.

Seed a well-order with the image of a supplied injection $\omega \to X$. Recursively for $\xi <h(X)$ append $a(Y_{\xi },{<}_{\xi })$, where $Y_{\xi }$ consists of the seed followed by previously appended elements in index order. At limits take the union of these extending well-orders. Each stage is an infinite well-ordered subset of $X$, so the rule always yields a fresh point. Thus the appended points inject $h(X)$ into $X$, a contradiction.