If $f$ is a uniformly continuous function show that $g = f(x) - f(y)$ is uniformly continuous on all of $\mathbb{R}^2$ Problem statement:

Let $f: \mathbb{R} \to \mathbb{R}$ be a uniformly continuous function on $\mathbb{R}$ and let $g: \mathbb{R}^2 \to \mathbb{R}$ be defined by $f(x) - f(y)$. Then $g$ is uniformly continuous on $\mathbb{R}^2$.

My attempt at a proof:
I think I am mostly getting confused about the notation.
Let $p = (x_p,y_p)$ and $q = (x_q,y_q)$ be values in $\mathbb{R}^2$. We wish to show that given $\epsilon > 0$, there exists a $\delta > 0$ for which $|g(p)-g(q)| < \epsilon$ for all $p,q \in \mathbb{R}^2$ with $d_{\mathbb{R}^2}(p,q)<\delta$.
We will use the fact that $g(x,y) = f(x) - f(y)$, where $f$ is uniformly continuous, and $x,y \in \mathbb{R}$.
Let $\epsilon>0$ be given. Then, $\exists$ a $\delta > 0$ for which $|f(x) - f(y)| < \epsilon$ for all $x,y \in \mathbb{R}$ with $|x-y| < \delta$, or $|g(x,y)| < \epsilon$.
Ugh I am getting seriously lost in the notation. Can anyone tell me the best way to notate this problem?
 A: Let $\epsilon>0$ be given. Since $f$ is uniformly continuous, there exists a $\delta>0$ such that $|f(x)-f(y)|<\frac{\epsilon}{2}$ whenever $|x-y|<\delta$. Now let $(x_1,y_1),(x_2,y_2)$ be two points in $\mathbb{R^2}$ such that $|(x_1,y_1)-(x_2,y_2)| <\delta$, then, in particular $|x_1-x_2|<\delta$ and $|y_1-y_2|<\delta$. Therefore,
\begin{align*}
|g(x_1,y_1)-g(x_2,y_2)| &= |f(x_1)-f(y_1) -f(x_2)+f(y_2)|\\
&\leq |f(x_1)-f(x_2)| +|f(y_1)-f(y_2)|\\
&<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.
\end{align*}
A: Given $\epsilon>0$ there is $\delta>0$ such that if $|x_1-x_2|<\delta$ then $|f(x_1)-f(x_2)|<\epsilon/2$.
Let $(x_1,y_1)$, and $(x_2,y_2)$ be such that $\|(x_1,y_1)-(x_2,y_2)\|<\delta$. Then $|x_1-x_2|<\delta$ and $|y_1-y_2|<\delta$. Therefore
$$\begin{align}|g(x_1,y_1)-g(x_2,y_2)|&=|f(x_1)-f(y_1)-f(x_2)+f(y_2)|\\&\leq|f(x_1)-f(x_2)|+|f(y_1)-f(y_2)|<\epsilon\end{align}$$
A: Let $\varepsilon>0$. Since $f$ is uniformly continuous, there is a $\delta>0$ such that for every $x,a \in \mathbb{R}$ we have:
$$
(|x-a|<\delta) \Longrightarrow (|f(x)-f(a)|<\frac\varepsilon2).
$$
Thus, for every $(a,b), (x,y) \in \mathbb{R}^2$ with $\max\{|x-a|,|x-b|\}<\delta$ we have:
\begin{eqnarray}
|g(x,y)-g(a,b)|&=&\left|f(x)-f(a)-(f(y)-f(b))\right|\\
&\le& |f(x)-f(a)|+|f(y)-f(b)|\\
&<&\frac\varepsilon2+\frac\varepsilon2=\varepsilon.
\end{eqnarray}
