Continuous functions of $(0,1)$ form a metric space

Note that $\rho(x,y)$ is essentially the Lebesgue measure of $E(x,y)$, and this can be used to elimiate the fuss of cutting up the intervals differently mentioned in @Calvin_Khor's comment.

We could bring a more general perspective into the metric.

Let $(\Omega, \mathcal F, P)$ be a probability space, and $(M, d)$ be a bounded metric space, then the set of measurable functions $f:\Omega\rightarrow M$ is equipped with the metric $$d(x,y):=\int_\Omega d(x(t), y(t))dt$$

The proof for this is pretty straightforward using $d$ satisfies the triangle inequality.

In your case, $\Omega = (0, 1)$ and $P$ is the Lebesgue measure. $M=\mathbb R$ equipped with the discrete metric $d(x,y)=1$ whenever $x\not=y$, and $\rho(x,y)$ is the probability of $x\not=y$.