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 $ε/2$. 

You can as well show that the interior of $R$ is empty in the standard topology. Since any open ball around every point of $R$ contains points in $R$ and outside $R$ then the boundary of $R$ is $R$. By definition, closure of is the union of $R$ and its boundary. Therefore Closure of $R$ is just $R$.
Please criticize this proof so I can make corrections if any.
