Convolution of Mollifier and function $f$ converges to $f$ on all compact subsets?

I am studying the book by Evans L.C., Gariepy R.F. - Measure theory and fine properties of functions. On the chapter about local approximation of Sobolev functions, he introduces the function $$\eta:\mathbb{R}^n\longrightarrow\mathbb{R}$$ by $$\eta(x)= \begin{cases} c\exp{\left(\frac{1}{{\lvert x\rvert}^2-1}\right)}, & \text{if} \ \lvert x\rvert <1, \\ 0 & \text{else}, \end{cases}$$ where the constant $$c$$ is chosen in a way, such that $$\int_{\mathbb{R}^n}\eta(x)dx=1$$ holds. Also he defines $$\eta_\varepsilon(x)=\frac{1}{\varepsilon^n}\eta\left(\frac{x}{\varepsilon}\right), \ \ \ \ \varepsilon>0,x\in\mathbb{R}^n,$$ and then lastly $$f^\varepsilon=\eta_\varepsilon\ast f,$$ that is $$f^\varepsilon(x)=\int_{\Omega}\eta_\varepsilon(x-y)f(y)dy, \ \ \ \ x\in \Omega_\varepsilon.$$ He then states that for $$f\in C(\Omega)$$ we have $$f^\varepsilon\longrightarrow f$$ uniformly on all compact subsets of $$\Omega$$.

Here is how the proof goes: Let $$A\subset\subset\Omega$$ arbitrary and chose $$A\subset B\subset\Omega$$. Then for every $$x\in A$$ we have $$f^{\varepsilon}(x)=\frac{1}{\varepsilon^n} \int_{\overline{B_\varepsilon}(x)} \eta\left(\frac{x-y}{\varepsilon}\right) f(y) d y=\int_{\overline{B_1}(0)} \eta(z) f(x-\varepsilon z) d z.$$ And since $$\int_{\overline{B_1}(0)}\eta(z)dz=1$$, it follows that $$\lvert f^\varepsilon(x)-f(x)\rvert\leq \int_{\overline{B_1}(0)} \eta(z) \lvert f(x-\varepsilon z)-f(x)\rvert d z.$$ If $$f$$ is uniformly continuous on $$B$$, we conclude from this estimate that $$f^\varepsilon\rightarrow f$$ uniformly on $$A$$.

My question is, why does the chose the ball around x as, why does he chose the $$A$$ and $$B$$ like that and why does the result follow from the estimate?

Assuming that $$\{x : d(A, x) \leq \varepsilon\} \subset B$$, Bound $$|f(x - \varepsilon z) - f(x)| \leq \sup_{y,z \in B, |y - z| \leq \varepsilon}|f(y) - f(z)| =: \omega_B(\varepsilon)$$ in the integral to get $$|f^{\varepsilon}(x) - f(x)| \leq \omega_B(\varepsilon)$$. Uniform continiuity of $$f$$ on $$B$$ means $$\lim_{\varepsilon \searrow 0}\omega_B(\varepsilon) = 0$$.