Prove that the $x$-axis in $\Bbb R^2$ with the Euclidean metric is closed 
I want to show that the $x$-axis is closed. 

Below is my attempt - I would appreciate any tips on to improve my proof or corrections:
Let $(X,d)$ be a metric space with the usual metric.
Want to Show: $\{(x,y) | x ∈ \Bbb R, y = 0\}$ is closed
Claim: $\{(x,y) | x ∈ \Bbb R, y ≠ 0\}$ is open
Proof: Let $\{(x,y) | x ∈ \Bbb R, y ≠ 0\} = C$. 
Let $z$ be an arbitrary $(a,b) ∈ C$ and let $$ε = \min\{d(z,(0,y)), d(z,(x,0))\}.$$
Then for any $p ∈ $B_ε$(z)$, $B(p) ∈ \{(x,y) | x ∈ \Bbb R, y ≠ 0\}$ with radius $ε/2$. 

Another fun way you might approach this problem is to let $f : \mathbb{R}^2 \to \mathbb{R}$ be defined by $f(x,y) = y^2$. If you know/can show that $f$ is continuous, then it will imply that $f^{-1}(\{0\})$ is closed.
