Does an homeorphism function maps an interval in X to an interval in Y?

Question

Let $(X,\leq_X)$ and $(Y,\leq_Y)$ be two well-ordered sets. Let $f:X\rightarrow Y$ be an homeorphism between $X$ and $Y$ and their order topology (relative to $\leq_X$ and $\leq_Y$). My question is, suppose I have $x_1, x_2\in X$, such that $x_1\leq_X x_2$, so $U=(x_1,x_2)$ is an interval in $X$ and the order topology relative to $\leq_X$, does this imply that $f[U]$ is an interval in $Y$ and the order topology relative to $\leq_Y$? since $f$ is homeorphism I know that $f[U]$ is an open set, but is it necessarily also an interval? If it is, does this mean $f$ is also an isomorphism?