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 $

Let $f(x,y) = \text{max}(x,y)$. 
To show continuity we can use the collection of open squares as the basis for the space $\Bbb R \times \Bbb R$.
Consider any point $(x_0,y_0) \in \Bbb R^2$ with $z_0 = f(x_0,y_0)$.
Let the challenge $\varepsilon \gt 0$ be given.
We need to find a $\delta \gt 0$ such that 
$\tag 1 f\big((x_0-\delta, x_0+\delta) \times (y_0-\delta, y_0+\delta)\big) \subset (z_0-\varepsilon, z_0+\varepsilon) $
Set $\delta = \varepsilon$; we show that
$\tag 2 (x,y) \in (x_0-\varepsilon, x_0+\varepsilon) \times (y_0-\varepsilon, y_0+\varepsilon) \text{ implies } f(x,y) \in  (z_0-\varepsilon, z_0+\varepsilon) $
Suppose $z_0 = x_0$ so that $y_0 \le x_0$. When $f(x,y) = x$ or $f(x,y) = y$   simple arguments show that $\text{(2)}$ is true (the case $f(x,y) = x$ is trivial).
A symmetric argument shows that $\text{(2)}$ is true whenever $z_0 = y_0$.
