 Solution

The space is not compact because it is not complete. For $n\ge 2$, define the continuous functionThese functions converge in the L1 norm to the step function $h=1_{(1/2,1]}$, because $f_n-h$ is supported on an interval of length $2/n$ and bounded in absolute value by one. They are consequently Cauchy for the metric $d$, since for sufficiently close pairs the outer minimum with $1$ does nothing.

If $(f_n)$ converged in this metric to some $f\in C[0,1]$, it would also converge to $f$ in $L^1$. Uniqueness of an $L^1$ limit would give $f=h$ almost everywhere. Continuity would then force $f=0$ on $[0,1/2)$ and $f=1$ on $(1/2,1]$, which is impossible at $1/2$. Thus this Cauchy sequence has no limit in $C[0,1]$, while every compact metric space is complete.