Explicit examples are moderately difficult to construct, but it’s not too hard to come up with non-constructive examples; here’s one such.
For $x,y\in\mathbb{R}$ define $x\sim y$ iff $x-y\in \mathbb{Q}$; it’s easy to check that $\sim$ is an equivalence relation on $\mathbb{R}$. For any $x\in\mathbb{R}$, $[x] = \{x+q:q\in\mathbb{Q}\}$, where $[x]$ is the $\sim$-equivalence class of $x$. In particular, each equivalence class is countable. For any infinite cardinal $\kappa$, the union of $\kappa$ pairwise disjoint countably infinite sets has cardinality $\kappa$, so there must be exactly as many equivalence classes as there are real numbers. Let $h$ be a bijection from $\mathbb{R}/\sim$, the set of equivalence classes, to $\mathbb{R}$. Finally, define $$f:(0,1)\to\mathbb{R}:x\mapsto h([x])\;.$$
I claim that if $V$ is any non-empty open subset of $(0,1)$, $f[V]=\mathbb{R}$, which of course ensures that $f$ is open. To see this, just observe that every open interval in $(0,1)$ intersects every equivalence class. (It should be no trouble at all to see that $f$ is wildly discontinuous!)
Let us define for a function $f\colon X \to Y$ between metric spaces the minimal nondecreasing modulus of continuity as
$$\omega_f(\delta) := \sup \{ d(f(x),f(y)) : d(x,y) \leqslant \delta\}.$$
Among all monotonic functions that are admissible as a modulus of continuity of $f$, that is the smallest one. Non-monotonic moduli of continuity can be smaller at some values of course, consider a periodic function for example.
$f$ is uniformly continuous if and only if
$$\lim_{\delta \searrow 0} \omega_f(\delta) = 0.$$
The function $\omega_f$ is subadditive, $\omega_f(\delta + \varepsilon) \leqslant \omega_f(\delta) + \omega_f(\varepsilon)$ for functions whose domain is a convex domain in $\mathbb{R}^n$, hence $\omega_f(t)$ is finite for all $t \in [0,\infty)$ if $\omega_f(\delta) < \infty$ for some $\delta > 0$ then. For a space with infinitely many connected components that are at a positive distance from each other, $\omega_f(t)$ can be infinite for finite $t$ even for uniformly continuous $f$. We have
$$\omega_{\omega_f} \leqslant \omega_f,$$
so if $f$ is uniformly continuous, so is $\omega_f$, for functions with good domain.
Thus the question in the title can be answered affirmative, a uniformly continuous function $\mathbb{R}\to \mathbb{R}$ always has a (uniformly continuous) modulus of continuity that attains only finite values on $[0,\infty)$.
Note that $\lim\limits_{t\searrow 0} \omega(t) = \omega(0) = 0$ is part of the definition of a modulus of continuity, so if a function $f$ admits a global modulus of continuity, it must be uniformly continuous, and then allowing moduli of continuity that attain the value $\infty$ for finite $t$ is somewhat pointless for functions on good domains, but for functions on spaces with separated components, it can be necessary. And when one considers only local moduli of continuity (at a fixed point $x_0$), allowing infinite values is a simple way to not care about the behaviour far enough away from $x_0$.
Best Answer
The answer to the question in the title is negative.
We know that if a function $f:I \to \Bbb{R}$ is continuous and injective then its inverse $f^{-1}:f(I) \to I$ is also continuous.
Here $I$ denotes an interval .
Now since $f$ is open and a bijection we have that $f^{-1}$ is a continuous function,so $(f^{-1})^{-1}$ is also continuous.