Let $\alpha,\beta$ be ordinals.
Suppose that there is an order embedding $f: \alpha\hookrightarrow\beta$.
Show $\alpha \le \beta$, i.e. $\alpha\subset \beta$.
Try 1
There are two possibilities: $\alpha\subset \beta$ or $\beta\subset\alpha$. In the first case, we are done. In the second case, we can consider the composition $\beta\hookrightarrow \alpha \overset{f}{\hookrightarrow}\beta$ were the first arrow is the obvious inclusion map. This would produce an embedding from $\beta$ to itself, but it need not be surjective, so it doesn't seem useful.
Try 2
I also tried to fix $\alpha$ and proceed by transfinite induction on $\beta$, but got stuck. For instance, if I assume
$$\alpha\hookrightarrow \beta \implies \alpha\subset\beta$$
and take an embedding $f:\alpha\hookrightarrow S(\beta)=\beta\cup\{\beta\}$, we can distinguish two cases: $\beta\not\in f(\alpha)$ or $\beta\in f(\alpha)$. In the first case, it follows that actually $f:\alpha\hookrightarrow\beta$, so we can apply the induction hypothesis. I don't see how to proceed with the other case.
Any hints?
HINT: The map $\beta\hookrightarrow \alpha \overset{f}{\hookrightarrow}\beta$ is extremely useful: if it existed, and $\beta<\alpha$, there would be an order-embedding from $\beta$ to a proper subset of $\beta$. Prove by induction on $\beta$ that this is impossible.