Show $\inf_f\int_0^1|f'(x)-f(x)|dx=1/e$ for continuously differentiable functions with $f(0)=0$, $f(1)=1$.

Since $$ \int_0^1(f'(x)-f(x))\,e^{-x}\,\mathrm{d}x=\frac{f(1)}e-f(0)\tag{1} $$ we have $$ \int_0^1(f'(x)-f(x))\,\mathrm{d}x-\left(\frac{f(1)}e-f(0)\right)=\int_0^1(f'(x)-f(x))\left(1-e^{-x}\right)\mathrm{d}x\tag{2} $$ To make $(2)$ as small as possible, we would like to concentrate $f'(x)-f(x)$ where $1-e^{-x}$ is small; that is, near $x=0$.

Consider $f_a(x)=x^ae^{x-1}$ for $a\gt0$. We get $f_a'(x)-f_a(x)=ax^{a-1}e^{x-1}$. $$ \begin{align} \lim_{a\to0^+}\int_0^1|f_a'(x)-f_a(x)|\,\mathrm{d}x &=\lim_{a\to0^+}\int_0^1ax^{a-1}e^{x-1}\,\mathrm{d}x\\ &=\lim_{a\to0^+}\left(\left[x^ae^{x-1}\vphantom{\int}\right]_0^1-\int_0^1x^ae^{x-1}\,\mathrm{d}x\right)\\ &=\lim_{a\to0^+}\left(1-\int_0^1x^ae^{x-1}\,\mathrm{d}x\right)\\ &=1-\int_0^1e^{x-1}\,\mathrm{d}x+\lim_{a\to0^+}\left(\int_0^1\left(1-x^a\right)e^{x-1}\,\mathrm{d}x\right)\\[6pt] &=e^{-1}\tag{3} \end{align} $$ Since $$ 0\le\int_0^1\left(1-x^a\right)e^{x-1}\,\mathrm{d}x\le\int_0^1\left(1-x^a\right)\,\mathrm{d}x=\frac{a}{1+a}\tag{4} $$ the Squeeze Theorem says that $$ \lim_{a\to0^+}\left(\int_0^1\left(1-x^a\right)e^{x-1}\,\mathrm{d}x\right)=0\tag{5} $$ Thus, $(3)$ implies the bound is sharp.

Added: In light of everything said, the easiest approach seems to be to just take $f_\epsilon(x) = \eta(x/\epsilon) e^{x-1}$, where $\eta$ is a smooth function with $\eta(0) = 0$ and $\eta(x) = 1$ for $x\geq 1$ (so that $\int_0^1 \eta'= 1$). The integral is then $e^{-1}\int_0^\epsilon \epsilon^{-1}\eta'(x/\epsilon)e^x\,dx = e^{-1}\int_0^1 \eta'(x) e^{\epsilon x}\,dx$, which tends to $e^{-1}$ as $\epsilon\to0$. I blame the desire to write things down in terms of elementary functions for the over-complication below.

Here's an idea for a sequence of minimizers. (As the comment points out your proof shows there's no actual minimizer.) Take $f_\alpha(x) = x^\alpha e^{x-1}$ and let $\alpha \to 0$; the point is to make the function look more and more like an exponential as the parameter is varied. Note that $f_\alpha'(x) - f_\alpha(x) = \alpha x^{\alpha - 1} e^{x-1}$, and making the substitution $y = x^\alpha$ in the integral should show that the integrals tend to $e^{-1}$ as $\alpha \to 0$.

Edit: John pointed out that $f_\alpha$ isn't differentiable at $0$. To fix this, take some smooth approximation to $f_\alpha$ instead, where the approximation becomes better (sufficiently fast) as the parameter tends to $0$. To be explicit, you could use $g_\alpha = \eta_\alpha f_\alpha$, where $\eta_\alpha$ is smooth and increasing with $\eta_\alpha(0) = 0$ and $\eta_\alpha(\epsilon) = 1$ for some small $\epsilon$ (depending on $\alpha$). For the approximation to work you'll just need to make sure that $\int_0^\epsilon \eta_\alpha'(x) x^\alpha \to 0$ with $\alpha$. You can arrange for $\eta_\alpha'\lesssim 1/\epsilon$, and then the integral will be $\lesssim\epsilon^\alpha$, so, for instance, $\epsilon = \epsilon(\alpha) = \alpha^{1/\alpha}$ should do.

Here is another method. Let $f \in C^1([0,1])$ with $f(0)=0$ and $f(1)=1$, and $g=f'-f$. then solving the ODE : $f' =f +g$ with $f(0)=0$ gives $$f(x) = e^x \int_0^x e^{-t} g(t) dt.$$ Since $f(1)=1$, we get $$ \int_0^1 e^{-t} g(t) dt = \frac{1}{e} \quad \quad (1). $$ By Holder we have : $| \int_0^1 e^{-t} g(t) dt| \leq \|g\|_1$. Hence $$\|g\|_1 = \int_0^1 |f'-f| \geq 1/e,$$ and equality holds iff $g$ goes to the dirac $1/e.\delta_0$ (with the condition (1)).