= Solution
The space is not compact because it is not <complete metric space>[complete]. For $n\geq2$, define the continuous function
$$
f_n(t)=\begin{cases}
0,&0\leq t\leq\frac12-\frac1n,\\
\frac n2\left(t-\frac12+\frac1n\right),&\frac12-\frac1n<t<\frac12+\frac1n,\\
1,&\frac12+\frac1n\leq t\leq1.
\end{cases}
$$
These functions converge in the <L1 norm> to the <step function> $h=\boldsymbol1_{(1/2,1]}$, because $f_n-h$ is supported on an interval of length $2/n$ and bounded in <absolute value function>[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.
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