Let $A$ be a chain and $B$ be a partially ordered set. Let $f: A \rightarrow B$ be a 1-1 function for which $a \leq b$ implies $f(a) \leq f(b)$. Prove $f$ is order-isomorphic.
Let $f: A \rightarrow B$.
Let $x_0, x_1 \in A, f(x_0), f(x_1) \in B$.
Then $f(x_0) = f(x_1) \Longrightarrow x_0 = x_1$. So $f$ is order-isomorphic under $=$.
Now let $f(x_0) < f(x_1)$, so $x_0 \neq x_1$. But $A$ is a chain, so either $x_0 < x_1$ or $x_1 < x_0$. Suppose $x_1 < x_0 \rightarrow f(x_0) < f(x_1)$. What now?
The key observation here is that $x_1<x_0$ implies $x_1\le x_0,$ which implies by hypothesis that $f(x_1)\le f(x_0).$ Hence, you have $f(x_1)\le f(x_0)<f(x_1),$ which cannot happen in a partially ordered set.