Here I think it is implict that we have to use the metric induced by the norm. Also, an isometry here means a surjective function that preserves distances between points.
Well, I tried to find a isometric embedding that is also surjective. But I couldn't, and I think that the function that solves my problem isn't to intuitive (so I won't find it too easily) and so I came here asking for it... (hope this isn't duplicated).
My bounded creativity suggested me only $\varphi(f)(x)=f(x)b +(1-f(x))a$, but I think $\varphi$ is not surjective (and it doesn't even preserves distances hehe :/)
Help me, please!