Metric space is open 
For any $n=1,2,...$ and any $a\in \mathbb{R}$, the subset of $E^n$
  given by $\{(x_1,...,x_n)\in E^n : x_1>a\}$ is open.

I don't know why it's open?

Hint: Given $x = (x_1, ..., x_n) \in E^n$, consider the (open) ball of radius $$\frac{|x_1 - a|}{2} > 0$$ around $x$. Do you see why the ball is contained in $E^n$?
