How to show that $C_0 (U)$ is a Banach space

Let $$\emptyset \neq U \subset \mathbb{R} ^d$$ be an open set.

Define $$C_0(U) := \left\{ f \in C(U) : \forall \epsilon > 0 \exists K \subset U , K \mbox{ compact and }\sup_{x \in U \setminus K} |f(x)| < \epsilon \right\}$$.

To show is that $$(C_0 (U) , \left \lVert . \right \rVert_{\infty} )$$ is a Banach space.

I have huge struggles showing that $$C_0 (U)$$ is complete. I know I need a Cauchy sequence which converges in that norm. It would be great if someone can help me understand what to do with that set.

It's a standard and not too difficult result that that if $$f_n$$ is a Cauchy sequence of continuous functions in the superemum norm, it converges with this norm to a continuous function.
So, it remains to show that the uniform limit of continuous functions small outside of sets $$K_n$$ is also small outside of a compact set. Indeed, let $$f_n\to f$$ uniformly. Then, for a given $$\epsilon$$, we can find an $$N(\epsilon)$$ large enough so that whenever $$n\geq N(\epsilon)$$, we have $$|f_n(x)-f(x)|<\epsilon/2$$ for any $$x\in U$$. For this given epsilon, we can also find a $$K_{N,\epsilon}$$ compact in $$U$$ such that $$|f_{N}(x)|<\epsilon/2$$ for any $$x\in U\setminus K_{N,\epsilon}$$. So, by the triangle inequality, we show that $$f$$ is small outside of this set as well; let $$x\in U\setminus K_{N,\epsilon}$$, $$||f(x)|-|f_n(x)||<\epsilon/2\implies |f(x)|<\epsilon/2+|f_n(x)|<\epsilon$$ And the claim is shown.
If $$f_n$$ is a Cauchy sequence with respect to the uniform norm then you already know that $$f_n\rightarrow f$$ where $$f$$ is some continuous function on $$U$$ (because the continuous functions with the uniform norm is a Banach space). It is left to show that $$f\in C_0(U)$$.
Let $$\varepsilon>0$$, since $$f_n\rightarrow f$$ you have for $$N$$ sufficiently such that $$\|f_n-f\|<\varepsilon/2$$ whenever $$n>N$$. Fix such $$n>N$$, since $$f_n\in C_0(U)$$ there exists $$K$$ such that $$\|f_n\|_{U\backslash K}<\varepsilon/2$$
By the triangle inequality $$\|f\|_{U \backslash K} \leq \|f_n-f\| +\|f_n\|_{U\backslash K} < \varepsilon$$
*I use $$\|f\|_{U \backslash K}$$ to denote $$\sup_{x\in U\backslash K} |f(x)|$$.