Thank you Davide and robjohn, I merged your answers into the following:
To compute the following: $$ || f||_{C/W} = || x||_{C/W} = \inf_{f \in W} \sup_{x \in [-1,1]} |x + f(x)|$$
Let $g(x) = x + f(x)$. Then
$$ \int_0^1 g(x) d \mu = \frac{1}{2}$$ and
$$ \int_{-1}^0 g(x) d \mu = - \frac{1}{2}$$
$$ \implies || f||_{C/W} = \inf_{g \in W^\prime} \sup_{x \in [-1,1]} |g(x)| = \inf_{g \in W^\prime} ||g ||_\infty$$
where $W^\prime = \{ g: [-1,1] \rightarrow R | \int_0^1 g(x) d \mu = \frac{1}{2}, \int_{-1}^0 g(x) d \mu = - \frac{1}{2}\}$
(i) $$ || g||_\infty = || g||_\infty \mu([0,1]) \geq \int_0^1 g d \mu = \frac{1}{2}$$
$$ \implies ||f||_{C/W} \geq \frac{1}{2}$$
(ii) Now we want a function in $W$ whose supremum (=maximum) is $\frac{1}{2}$. That's easy:
$$ f(x) := -x - \frac{1}{2}$$ for $x \in [-1,0]$
$$ f(x) := x - \frac{1}{2}$$ for $x \in [0,1]$
$$ \implies || f ||_{C/W} = \frac{1}{2}$$