I have seen a proof of the statement, and its usually by transfinite induction. And I'm trying to find out why my proof doesn't work, it seems too simple:
Let $X$ be a well-ordered set. Define $X^{<x} = \{w \in X: w < x\}$ for all $x \in X$. Now define also $Z = \{X^{<x} : x \in X\}$. It is clear that $Z$ is an ordinal and there's a bijection that preserves order between $X$ and $Z$, namely $x \mapsto X^{<x}$.
This bijection is furthermore unique because we know that two ordinals admit an order-preserving bijection between them if and only if they are identical, and in that case the only order-preserving bijection is the identity map.
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Any feedback would be appreciated, thanks!
You may be on the right track, but with some current errors as pointed out in the comments. The idea of looking at $X^{<x}$ is good, but $Z$ is not necessarily an ordinal as it is currently defined.
For ease of speaking about it, it may help to give the desired order-preserving bijection a name, such as $\pi$. The goal is to have $X$ be the domain of $\pi$, and some ordinal be the range. For $x\in X$, define $\pi(x)=\{\pi(y)\mid y<x\}$, you may also think of this as the image of $X^{<x}$ under $\pi$ if you like to think in terms of $X^{<x}$. Then, $\pi$ has to have an ordinal as its range, i.e. the range of $\pi$ is transitive and well-ordered by $\in$ (linearly ordered by $\in$ and $\in$ is well-founded on $X$).
The following comes straight from the definition of $\pi$ and will be very useful:
$$\begin{align}\textrm{For }x,y\in X\textrm{, }x<y\iff\pi(x)\in\pi(y)\end{align}$$
The relation $<$ is transitive, so for any $x,y,z\in X$, $\pi(x)\in\pi(y)$ and $\pi(y)\in\pi(z)$ implies $\pi(x)\in\pi(z)$, so $\in$ is transitive on the range of $\pi$. Since $<$ is a linear ordering, $\in$ is a linear ordering on $X$. Finally, $\in$ must be well-founded on $X$ because any infinite decreasing $\ldots\in\pi(x_2)\in\pi(x_1)\in\pi(x_0)$ would correspond to an infinite decreasing chain $\ldots<x_2<x_1<x_0$ in $X$, which contradicts well-orderednss of $<$. (A shorter argument relying on the axiom of foundation is that $\in$ must be well-founded on $X$ directly from the axiom.)