A closed subspace of smooth functions in $L^1(0,1)$ is finite-dimensional The user "Q. So" asked that question and then deleted it, shortly after I wrote my answer. Since I think this is a nice problem and I also spent quite some time on it, I put both question and answer here for possible future reference.
Claim. Let $X\subset C^1[0,1]$ be a subspace that is closed in $L^1(0,1)$. Then $\dim X < \infty$.
 A: We endow $X$ with the $L^1$-norm. Let us show that the operator $T : X\to L^2(0,1)$, $Tf:= f'$, is bounded. For this, it suffices to show that $T$ is closable (because then $\overline T = T$ is closed and so $T$ is bounded by the closed graph theorem). Ok, so let $f_n\in X$ be such that $f_n\to 0$ and $\|f_n' - g\|_2\to 0$ with some $g\in L^2(0,1)$. Then, for any $\varphi\in C_c^\infty(0,1)$ we have
\begin{align*}
\|\varphi\|_\infty\|f_n'-g\|_2
&\ge\|(f_n'-g)\varphi\|_2\ge\|(f_n'-g)\varphi\|_1\\
&\ge\left|\int_0^1(f_n'-g)\varphi\,dt\right| = \left|\int_0^1f_n\varphi'\,dt + \int_0^1g\varphi\,dt\right|.
\end{align*}
Now, as
$$
\left|\int_0^1 f_n\varphi'\,dt\right|\le \|\varphi'\|_\infty\|f_n\|_1\to 0,
$$
it follows that $\int_0^1 g\varphi\,dt=0$ for all $\varphi\in C_c^\infty(0,1)$. Hence, $g=0$. This proves that $T$ is bounded.
Now, consider the space $X_0 := \{f\in X : f(0) = 0\}$. Since $X_0$ is $1$-codimensional in $X$ (it is the kernel of a linear functional), it suffices to prove that $\dim X_0 < \infty$. For $f\in X_0$ and $x\in [0,1]$ we have
$$
|f(x)| = \left|\int_0^x f'\,dt\right|\le\int_0^1|f'|\,dt = \|Tf\|_1\le\|Tf\|_2\le \|T\|\|f\|_1.
$$
Thus, $\|\cdot\|_\infty$ and $\|\cdot\|_1$ are equivalent norms on $X_0$. Therefore, we only have to show that $B_{X_0}$ (the closed unit ball of $X_0$) is equicontinuous (Arzela-Ascoli). But this follows from
$$
|f(x)-f(y)|\le\int_x^y|f'|\,dt\le\sqrt{|x-y|}\|f'\|_2\le\sqrt{|x-y|}\|T\|\|f\|_1 = \|T\|\sqrt{|x-y|}.
$$
