Define $T: L^1 [0,1] \rightarrow C[0,1] $ as $$Tf(x) = \int_0^x f(t)dt$$ In my understanding it's compact since $$sup | \int_0^x f_n(t)dt - \int_0^xf_m(t)dt | \leq \int_0^x |f_n(t) -f_m(t)|dt \leq \int_0^1 |f_n(t) -f_m(t)|dt = \Vert f_n-f_m\Vert_1$$ So for any convergence sequence $\{f_n\} \in L{[0,1]}$, $\{Tf_n\} $ is a convergence sequence in $ C{[0,1]}$.
But the answer is that the operator is noncompact. What goes wrong?
On
No, this operator is not compact. Let $f_n=nI_{(0,\frac 1n)}$. Then $(f_n)$ is bounded in $L^{1}[0,1]$. Now $Tf_n (x) = 1$ for $x >\frac 1 n$ and $Tf_n(0)=0$. This implies that $(Tf_n)$ conveges pointwise to the discontinuous function $f$ defined by $f(x)=1$ for $x >0$ and $f(0)=0$. Hence $T(f_n)$ cannot have any subsequence converging uniformly.
This is known as the Volterra operator, and indeed it is compact. It is an integral operator of the form $$ Tf(x) = \int_0^1 K(x,t)f(t)\ \mathsf dt $$ with kernel $K(x,t) = \mathsf1_{(0,x)}(t)$. Now by Tonelli's theorem we have $$ \int_{[0,1]^2} |K(x,t)|^2\ \mathsf d(x\times t) = \int_0^1\int_0^x \ \mathsf dt\ \mathsf dx = \frac12<\infty, $$ so that $K\in L^2([0,1]^2)$. It follows that $T$ is a Hilbert-Schmidt operator and thus is compact, as an operator from $L^2([0,1])$ to $L^2([0,1])$.
However, as pointed out by @Kabo Murphy's answer, $T$ is not compact considered as an operator from $L^1([0,1])$ to $C([0,1])$.