Please check my proof
I think the my proof is wrong for prove continuous at every point,because I don't understand about uniformly convergence
$$|x-x_{0}|<\delta\rightarrow |f(x)-f(y)|<\epsilon $$
$$ \rightarrow ||x|-|y||<\epsilon $$
$$\rightarrow |x-y|<\epsilon $$
Choose $\delta =\epsilon $ therefore |x| is continuous at every point in R
A function $f(\cdot)$ is continuous at $x_0$ if for every $\epsilon > 0$, there exists a $\delta$ such that for any point $x_0$, $|x - x_0| < \delta \implies |f(x) - f(x_0)| < \epsilon$. In general, this $\delta$ is a function of both $\epsilon$ and $x_0$. (We could use $x$ and $y$ instead of $x_0$ and $x$, but in your question you seem to have switched conventions.) If such a $\delta$ exists for all $\epsilon > 0$ and all $x$ in the domain of $f$, then we say the function is continuous everywhere.
If for every $\epsilon$, the same $\delta$ will work for all $x$, then we say the function is uniformly continuous. Continuity everywhere does not imply uniform continuity, and uniform continuity over a subset of the domain does not imply continuity everywhere. It's unclear exactly what you are asked to prove, but in this case we can prove that |x| is uniformly continuous everywhere.
Note that if $x_0$ and $x$ are both non-positive or both non-negative, then $||x_0| - |x|| = |x - x_0|$. If they have opposite sign, then $||x_0| - |x|| \leq |x - x_0|$.
So, your intuition that for any $\epsilon$ we should just pick $\delta = \epsilon$ was correct. $|x - x_0| < \delta$ implies $|x - x_0| < \epsilon$, which implies:
$|f(x_0) - f(x)| = ||x| - |x_0|| \leq |x - x_0| < \epsilon$
It seems you have the right idea, but are struggling a bit with notation and terminology.