If $f:\kappa \to \kappa$ is strictly increasing and continuous, why then holds that
for every $\alpha < \kappa$, $f(\alpha) \geq \alpha$
Furthermore, why in this case also holds that $f[\kappa]$ is closed in $\kappa$? Where by closed I mean the following definiton:
A subset $C \subset \kappa$ is said to be closed in $\kappa$ if $\forall\lambda<\kappa(\lambda$ is a limit ordinal $\wedge$ $C\cap\lambda$ is unbounded in $\lambda \to \lambda \in C$)
The following solution emerged in the comments:
To show that $f(\alpha)\ge\alpha$, we use the fact that $f$ is increasing (continuity is not needed here). Let, for contradiction, $\beta$ be the least ordinal such that $f(\beta)<\beta$. Then for each $\gamma<\beta$, we have $f(\gamma)\ge \gamma$; so since $f$ is increasing, we need to have $f(\alpha)>\gamma$ for all $\gamma<\alpha$. So $f(\alpha)\ge\alpha$.
To show that $f[\kappa]$ is closed, we use the continuity of $f$ (increasingness is not needed here). Let $\lambda<\kappa$ be such that $f[\kappa]\cap\lambda$ is unbounded in $\lambda$; then let $S=\{\eta: f(\eta)<\lambda\}$ be the preimages of the values of $f$ below $\kappa$. We now pick the least $\beta$ such that $S\cap \beta$ is unbounded in $\lambda$ (this trick was not mentioned in the comments - do you see why it's necessary? - but we can get around it if you want by using the fact that $f$ is increasing); it's not hard to show that continuity then forces $f(\beta)=\lambda$.