Proof that $\max(x_1,x_2)$ is continuous I haven't done a proof in years and I've become a little stuck on this, I'd appreciate it if somebody could tell me if I've approached the problem correctly...

Question: Prove that the following function $f(x_1; x_2) = \max[x_1; x_2]$, $x_1, x_2 \in \Bbb R$ is continuous over $\Bbb R^2$ ($\max [x_1; x_2] = x_1$ if $x_1 \geq x_2$). Use definition of continuity.

Function $f\colon \mathbb{R}^2 \to \mathbb{R}$ is continuous at $x_0 \in \mathbb{R}$ if $\forall \epsilon > 0 \exists \delta >0 :\mbox{ if }|x-x_0| <\delta \implies |f(x_0) - f(x)|<\epsilon$
Proof: Consider $(x_1,x_2) \in \mathbb{R}^2$ and $x_1>x_2$. Then  $\forall \epsilon > 0$ choose $\delta > 0 : ||(x_1,x_2) - (x_0,x_0)|| < \delta = \epsilon $, then;
$|\max(x_1,x_2) - \max(x_0,x_0)| = |x_1 - x_0| \le ||(x_1,x_2) - (x_0,x_0)|| < \delta = \epsilon $

If $x_2>x_1$ then $\max(x_1,x_2)=x_2$ and then it is continuous in that region. Similar analysis in the region $x_1>x_2$.
If $x_2=x_1$ then note that $|\max(y_1,y_2)-\max(x_1,x_2)|\le \max(|y_1-x_1|,|y_2-x_2|)$. 
Given $\varepsilon>0$, choose $\delta=\varepsilon$. If $||(y_1,y_2)-(x_1,x_2) ||<\delta$ then we have
$$
|\max(y_1,y_2)-\max(x_1,x_2)|\le \max(|y_1-x_1|,|y_2-x_2|)\le \sqrt{|y_1-x_1|^2+|y_2-x_2|^2}<\delta =\varepsilon. 
$$
