Let $\omega_1$ be the first uncountable ordinal. Look at the set $X=[0,\omega_1[$, the set of all ordinals $<\omega_1$. It is a well-ordered set with a smallest element. Since any ordinal strictly less than $\omega_1$ is countable, we can see that any sequence of points of $X$ must have a majorant.
Now with that in mind, let $f:X \to X$ be some map such that, for any $x$ big enough, we have $f(x)<x$. I need to show that there is some $c \in X$ such that, for any $x$ big enough, $f(x) \leq c$. A clue is provided in the exercise: suppose there is no such $c$, then we can construct a sequence $(z_n)$ of points of $X$ with the property that $z_{n+1}$ is the smallest $z$ in $X$ such that $f(x) \geq z_n$ for any $x \geq z$. Unfortunately I'm not able to see how this clue is helpful.
(Here is how this fact is related to non-paracompactness. Note that $X$, when endowed with the order topology, is locally compact Hausdorff but not compact. So pick some open covering $(U_\alpha)$ with no finite subcovering. We prove that $(U_\alpha)$ has no locally finite open refinement. Assume that $(V_\beta)$ is such a locally finite open refinement. Any $x$ belongs to some $V_{\beta(x)}$. There is some $f(x)<x$ such that $]f(x),x]$ is a subset of $V_{\beta(x)}$ (unless maybe $x$ is the smallest element, we discard this obvious case). But we apply the previous fact to $f$ and conclude that in fact our refinement can't be locally finite, using the obvious fact that $(V_\beta)$ is infinite. )
Thank you for your help.
(I may add that I'm able to prove that $X$ is not paracompact, in a maybe more straight forward way, but I still would like to understand the previously stated fact.)
Let $f:\omega_1\to\omega_1$ and $\alpha<\omega_1$ be such that $f(\xi)<\xi$ whenever $\alpha\le\xi<\omega_1$. Suppose that for each $\gamma<\omega_1$ the set $\{\xi<\omega_1:f(\xi)\le\gamma\}$ is bounded in $\omega_1$. Let $\beta_0=\alpha$. Given $\beta_n<\omega_1$, let
$$\beta_{n+1}=\min\{\eta<\omega_1:f(\xi)>\beta_n\text{ whenever }\eta\le\xi<\omega_1\}\;;$$
This exists by virtue of the assumption that $\{\xi<\omega_1:f(\xi)\le\beta_n\}$ is bounded.
Now let $\beta=\sup\{\beta_n:n\in\omega\}$. Then $\beta_{n+1}\le\beta$ for each $n\in\omega$, so $f(\beta)>\beta_n$ for each $n\in\omega$, and therefore $f(\beta)\ge\sup_{n\in\omega}\beta_n=\beta$. On the other hand, $\beta\ge\beta_0=\alpha$, so $f(\beta)<\beta$. This contradiction shows that there must be some $\gamma<\omega_1$ such that the set $\{\xi<\omega_1:f(\xi)\le\gamma\}$ is unbounded. In other words, for each $\eta<\omega_1$ there is a $\xi\in(\eta,\omega_1)$ such that $f(\xi)\le\gamma$.
Note that you misstated the conclusion: we can get a $\gamma$ so that $\{\xi<\omega_1:f(\xi)\le\gamma\}$ is unbounded, but we cannot necessarily get a $\gamma$ such that $f(\xi)\le\gamma$ for all sufficiently large $\xi$. To see this, let $f(\xi)=0$ whenever $\xi$ is a limit ordinal, and let $f(\xi+1)=\xi$ for each successor ordinal $\alpha+1$. Then $\{\xi<\omega_1:f(\xi)=0\}$ is unbounded, but for each $\alpha<\omega_1$ we have $f(\alpha+2)=\alpha+1>\alpha$.
This result is a special case of the pressing-down lemma, also known as Fodor’s lemma. The pressing-down lemma is an extremely useful tool when dealing with regular cardinals.