Obscure proof for $f\in \mathcal C_0(\mathbb R)\implies U_f\in u_f$ continuous. Let $$\mathcal C_b(\mathbb R)=\{f:\mathbb R\longrightarrow \mathbb R: f\ \text{continuous and bounded}\}\quad \text{and}\quad \mathcal C_0(\mathbb R)=\{f\in\mathcal C_b(\mathbb R)\mid \lim_{|x|\to\infty }f(x)=0\}$$
with the norm $\|f\|=\sup_{x\in\mathbb R}|f(x)|$. Let $f\in \mathcal C_b(\mathbb R)$ and let define $u_f:\mathbb R\longrightarrow \mathcal C_b(\mathbb R)$ defined by $u_f(y)(x)=f(x+y)$. Prove that $f\in\mathcal C_0(\mathbb R)$ implies $u_f$ continuous.
Proof
Let $f\in\mathcal C_0(\mathbb R)$ and $y\in\mathbb R$ fixed. We have to show that $$\lim_{z\to y}\|u_f(z)-u_f(z)\|=\lim_{z\to y}\sup_x|f(x+y)-f(x+z)|=0.$$
We can see that $u_f$ is continuous if and only if $f$ is uniformly continuous. 
Q1) I don't understand why ? In what the fact that $\sup_x|f(x+y)-f(x+z)|\underset{z\to y}{\longrightarrow }0$ will gives that $$\forall \varepsilon>0, \exists \delta>0: |z-y|<\delta\implies |f(z)-f(y)|<\varepsilon\ \ ?$$
Let $\varepsilon>0$ and let show that if $f\in \mathcal C_0(\mathbb R)$, the function is uniformly continuous. By definition, there is a $M$ s.t. $|f(t)|<\varepsilon$ if $t\notin [-M,M]$. We can suppose WLOG $|z-y|<1$, and thus if $|x+y|>M+1$ then $|x+z|>M$.
Q2) Why do we consider $|x+y|>M+1$ and it what it implies that $|x+z|>M$ ? I don't see the logic here
Therefore $|f(x+y)-f(x+y)|<2\varepsilon$ if $x\notin K=[M-1-y,M+1-y]$.
Q3) Since $|x+y|>M+1$ and $|x+z|>M$, we automatically have $|f(x+y)-f(x+y)|<2\varepsilon$, don't we ? So why taking $x\notin K=[M-1-y,M+1-y]$ ?
But $K+y=[-M-1,M+1]$ is compact, and thus the function is uniformly continuous on $K+y$. Let $\delta>0$ s.t. for all $t,w\in K+y$, $|t-w|<\delta\implies |f(t)-f(w)|<\varepsilon$. In other word, if $|y-z|<\delta$ then $|f(x+z)-f(x+y)|<\varepsilon$ for all $x\in K$.
Q4) I don't see how from "Let $\delta>0$ s.t. for all $t,w\in K+y$, $|t-w|<\delta\implies |f(t)-f(w)|<\varepsilon$" we get "if $|y-z|<\delta$ then $|f(x+z)-f(x+y)|<\varepsilon$ for all $x\in K$"
For the rest, it's fine. Thank you for your help.
 A: Let $\varepsilon>0$. Since $\lim_{\lvert x\rvert\to\infty}\lvert\,f(x)\rvert=0,$
there exists an $M>0$, such that
$$
\lvert x\rvert>M\quad\Longrightarrow\quad 
\lvert\, f(x)\rvert<\frac{\varepsilon}{3}.
$$
But as $f$ is continuous in $[-3M,3M]$, then is also uniformly continuous, and hence there exists a $\delta>0$, which is chosen to be less that $M$, such that
$$
\lvert y-z\rvert<\delta \quad\Longrightarrow\quad 
\lvert\, f(y)-f(z)\rvert<\frac{\varepsilon}{3}.
$$
Altogether, if $\lvert y-z\rvert<\delta$, then if $x+y\in [-2M,2M]$, then $x+z\in [-3M,3M]$, and hence
$$
\lvert\, f(x+y)-f(x+z)\rvert<\frac{\varepsilon}{3},
$$
due to uniform continuity in $[-3M,3M]$. If $\lvert x+y\rvert>2M$, then $\lvert x+z\rvert>M$, and hence
$$
\lvert\, f(x+y)-f(x+z)\rvert\le\lvert\, f(x+y)\rvert+\lvert\, f(x+z)\rvert <\frac{2\varepsilon}{3}.
$$
So, we have that
$$
\lvert y-z\rvert<\delta \quad\Longrightarrow\quad 
\lvert\, f(x+y)-f(x+z)\rvert<\frac{2\varepsilon}{3}
$$
for all $x\in\mathbb R$, and hence
$$
\lvert y-z\rvert<\delta \quad\Longrightarrow\quad 
\sup_{x\in\mathbb R}\lvert\, f(x+y)-f(x+z)\rvert\le\frac{2\varepsilon}{3}<
\varepsilon.
$$
