Let there be a series of continuous functions $u_n\colon \Omega\subseteq\mathbb{R}^n\to\mathbb{R}$ with nearly compact support, that is for $\varepsilon>0$ the set $u_n^{-1}(\mathbb{R}\setminus(-\varepsilon,\varepsilon))$ is compact, and let that series be uniformly convergent to $u$ according to the sup-norm.
So, continuity (thereby closedness for the above pre-image) for $u$ transfers by uniform convergence, but how do I show, technically, the boundedness of the above pre-image for $u$?
Why is there a unit in the ring those continuous functions with nearly compact support form if, and only if, $\Omega$ is compact?
Let $\epsilon>0$. For large enough $n$, $\|u_n-u\|<\frac \epsilon2$. $$|u(x)|\ge \epsilon \implies |u_n(x)|\ge \frac \epsilon 2,$$ i.e., $$u^{-1}(\Bbb R\setminus (-\epsilon,\epsilon))\subseteq u_n^{-1}(\Bbb R\setminus (-\tfrac\epsilon2,\tfrac\epsilon2))$$ and the latter set is compact.
Note that the function given by $f(x)=\frac1{1+\|x\|}$ is nearly compactly supported (and continuous). A unit element $e$ in your ring must have the property that $e(x)f(x)=f(x)$ for all $x\in\Omega$, hence $e(x)=1$ for all $x\in \Omega$. With $\epsilon:=1$, we have $e^{-1}(\Bbb R\setminus(-\epsilon,\epsilon)=\Omega$, and this must be compact if $e$ is nearly compactly supported.