This problem reminds me the classical Bolza example (named after Oskar Bolza) in the CoV, because $\inf F =0$ can be proved using sawtooth functions.
Obviously $F[u]\geq 0$, thus $\inf F \geq 0$. Now, take: $\phi (x):=\begin{cases} 1 -|x-1| &\text{, if } |x-1|\leq 1 \\ 0 &\text{, otherwise}\end{cases}$ and, for $N\in \mathbb{N}$, define $u_N:[0,1]\to \mathbb{R}$ via: $u_N(x):= \frac{1}{2N}\ \sum_{n=0}^{N-1}\phi (2Nx-2n)\; ;$ then $u_N$ is a nonnegative continuous function mapping $[0,1]$ onto $[0,1/(2N)]$ which is a.e. differentiable and, in particular: $u_N^\prime (x)=\pm 1\quad \text{for a.e. } x\in [0,1]\; ;$ hence: $\min \Big\{(u_N^\prime (x)-1)^2,(u_N^\prime (x)+1)^2\Big\} =0 \quad \text{for a.e. } x\in [0,1]$ and therefore: $F[u_N]=\int_0^1 u^2 +\int_0^1\min \Big\{(u_N^\prime (x)-1)^2,(u_N^\prime (x)+1)^2\Big\} \leq \frac{1}{4N^2}\; .$ The latter inequality yields $\inf F =0$.
For the continuity, you can argue as follows. You notice that using the reverse triangular inequality and other elementary tricks you get: $\begin{split} |F[u]-F[v]| &\leq \Big| \lVert u\rVert_2^2 -\lVert v\rVert_2^2\Big|\\ &\phantom{\leq} + \left| \int_0^1 \bigg( \min \Big\{ |u^\prime -1|,|u^\prime +1|\Big\}\bigg)^2 -\int_0^1\bigg( \min \Big\{ |v^\prime -1|,|v^\prime +1|\Big\}\bigg)^2\right|\\ &= (\lVert u\rVert_2+\lVert v\rVert_2)\ \Big| \lVert u\rVert_2 -\lVert v\rVert_2 \Big|\\ &\phantom{\leq} +\left| \left\lVert \min \Big\{ |u^\prime -1|,|u^\prime +1|\Big\}\right\rVert_2^2 -\left\lVert \min \Big\{ |v^\prime -1|,|v^\prime +1|\Big\} \right\rVert_2^2\right|\\ &\leq (\lVert u\rVert_2+\lVert v\rVert_2)\ \Big| \lVert u\rVert_2 -\lVert v\rVert_2 \Big|\\ &\phantom{\leq} +\left( \left\lVert \min \Big\{ |u^\prime -1|,|u^\prime +1|\Big\}\right\rVert_2 +\left\lVert \min \Big\{ |v^\prime -1|,|v^\prime +1|\Big\}\right\rVert_2 \right)\times \\ &\phantom{\leq +}\times \left| \left\lVert \min \Big\{ |u^\prime -1|,|u^\prime +1|\Big\}\right\rVert_2 -\left\lVert \min \Big\{ |v^\prime -1|,|v^\prime +1|\Big\}\right\rVert_2 \right|\\ &\leq (\lVert u\rVert_2+\lVert v\rVert_2)\ \lVert u-v\rVert_2\\ &\phantom{\leq} +\left( \left\lVert \min \Big\{ |u^\prime -1|,|u^\prime +1|\Big\}\right\rVert_2 +\left\lVert \min \Big\{ |v^\prime -1|,|v^\prime +1|\Big\}\right\rVert_2 \right)\times \\ &\phantom{\leq +} \times \left\lVert \min \Big\{ |u^\prime -1|,|u^\prime +1|\Big\} -\min \Big\{ |v^\prime -1|,|v^\prime +1|\Big\}\right\rVert_2\end{split}$ for each $u,v\in W^{1,2}(0,1)$. Now, it is easy to prove that function: $\mathbb{R}\ni s\mapsto \min \Big\{ |s-1|,|s+1| \Big\} \in \mathbb{R}$ is Lipschitz with constant $L=1$ (draw a picture), hence: $\tag{1} \begin{split} |F[u]-F[v]| &\leq (\lVert u\rVert_2+\lVert v\rVert_2)\ \lVert u-v\rVert_2\\ &\phantom{\leq} +\left( \left\lVert \min \Big\{ |u^\prime -1|,|u^\prime +1|\Big\}\right\rVert_2 +\left\lVert \min \Big\{ |v^\prime -1|,|v^\prime +1|\Big\}\right\rVert_2 \right)\ \lVert u^\prime -v^\prime\rVert_2 \; . \end{split}$ Finally, if you fix $v$ and let $u\stackrel{W^{1,2}}{\longrightarrow} v$ in (1) you obtain $|F[u]-F[v]|\to 0$ as you wanted.