Jonas: What you need to show is that if $Y$ is well-ordered, then any subset $Z$ of $Y$ is isomorphic to an initial segment of $Y$. How would you define such an isomorphism $f$? You do not have many possibilities: The least element of $Z$ must be mapped by $f$ to the smallest element of $Y$. The next element of $Z$ must be mapped by $f$ to the next element of $Y$, and so on. You just need to check that this works.
In more detail, say that $g$ is an approximation iff its domain is an initial segment of $Z$, its range is an initial segment of $Y$, and $g$ is an order isomorphism from its domain onto its range.
One checks (by considering least disagreements) that any two approximations coincide on the common part of their domains, so we can let $f$ be the common function resulting from pasting all the approximations together.
We just need to see that $f$ exhausts $Z$. For this, one can check that $f(x)\le x$ for all $x$, so one "does not run out of room."
There is a slightly more elegant (and formal) way of presenting this argument, by appealing to the recursion theorem.
For $x$ in $Z$, define $f(x)=\sup_Y({\rm succ}_Y f(y)\mid y\in Z, y\lt x)$, where ${\rm succ}_Y a$ means the successor in $Y$ of $a$. Then $f$ is precisely the map we wanted. (One still needs a little argument to see that it works).
A trickier question for another day is whether the recursion theorem is needed for the argument. This may be formalized by asking whether Zermelo set theory proves the result (so we do not have access to replacement), but there may be finer formulations.
(The above was written quickly. Let me know of infelicities or plain falsehoods and I'll try to edit.)