Recall that for all $f \in C(X)$, $f$ is uniformly continuous because $X$ is compact.
Given $F \subset C(X)$ a finite set, we show that it is equicontinuous. Fix $\varepsilon > 0$. We need to find som $\delta > 0$ such that some condition is satisfied.
For all $f \in F$, by uniform continuity of $f$, there exist $\delta_f$ such that etcetera. Your required $\delta$ is $\delta = \min_{f \in F} \delta_f$.
In particular, using the same argument, you can see that a finite union of equicontinuous sets is equicontinuous.
The family is not uniformly equicontinuous on $(1, +\infty)$. If it were, since each $f_a$ has a continuous extension to $[1, +\infty)$, the family of extensions would be uniformly equicontinuous on $[1, +\infty)$. (For every $\varepsilon > 0$ the same $\delta$ would work on $[1, +\infty)$ as on $(1, +\infty)$.) A fortiori we would obtain a uniformly equicontinuous family on $[1,2]$ by restriction, and the Ascoli-Arzelà theorem says every sequence in that family would have a uniformly convergent subsequence. But we have
$$\lim_{n \to \infty} f_n(x) = \begin{cases} 1 &\text{if } x = 1,\\ 0 &\text{if } x > 1, \end{cases}$$
thus a sequence whose pointwise limit is discontinuous. It follows that the sequence has no uniformly convergent subsequence, hence the family is not equicontinuous on $[1,2]$. (Note: on a compact domain the concepts of uniform equicontinuity and pointwise equicontinuity coincide.)
As for pointwise equicontinuity, we can show that the family is pointwise equicontinuous on $(1, +\infty)$ using the Ascoli-Arzelà theorem.
Consider a sequence $(f_{a_n})_{n \in \mathbb{N}}$ and the sequence $(a_n)$ of indices. Either $(a_n)$ has a convergent subsequence - in which case we may assume the whole sequence converges, say to $\alpha \in [0, +\infty)$ - or we have $a_n \to +\infty$.
Consider any compact interval $[u,v] \subset (1, +\infty)$. In the former case, $(f_{a_n})$ converges to $f_{\alpha}$ uniformly on $[u,v]$, in the latter case it converges uniformly to $0$ on $[u,v]$.
Thus by the Ascoli-Arzelà theorem the family of the restrictions $\{ f_a\rvert_{[u,v]} : a \in \mathbb{R}^+\}$ is equicontinuous, and that implies the original family is equicontinuous at all $x \in (u,v)$. Since every $x \in (1, +\infty)$ lies in some $(u,v)$ with $1 < u < v < +\infty$, the family is pointwise equicontinuous on $(1, +\infty)$.
Best Answer
Indeed, the idea of approximating $\Omega$ by a sequence of compact sets will work. Let $$ K_j:=\left\{x\in\mathbb R^n, \lVert x\rVert\leqslant j\right\}\cap \left\{x\in\mathbb R^n, d\left(x,\mathbb R^n\setminus\Omega \right)\leqslant 1/j\right\}, $$ where $d(x,S)=\inf\{\lVert x-y\rVert,y\in S\}$ and $\lVert \cdot\rVert$ denotes the Euclidean norm. Then $K_j$ is compact for each $j$ and $\Omega=\bigcup_{j\geqslant 1}K_j$. Moreover, notice that each compact subset of $\Omega$ is contained in some $\Omega_j$.
Now you can use a diagonal extraction process to find a subsequence for which the uniform convergence holds for each $K_j$ and in view of the previous remark, for each compact subset of $\Omega$.