Showing a subset of $C([0,1])$ is compact.

According to Arzelà–Ascoli theorem you only have to show that $\mathcal F$ is

• equicontinuous, i.e. $(\forall x\in[0,1])(\forall \varepsilon>0)(\exists \delta>0)(\forall f\in\mathcal F) (\forall y) |y-x|<\delta \Rightarrow |f(y)-f(x)|<\varepsilon$;
• pointwise bounded;
• closed in $C[0,1]$.

Both equicontinuity and pointwise boundedness follow from the Lipschitz condition $|f(x)-f(y)|\le |x-y|$.

To show pointwise boundedness you can notice that $|f(x)-f(0)|\le |x|=x$, which means $$f(0)-x \le f(x) \le f(0)+x.$$ If you apply integral $\int_0^1$ to the left inequality, you get $f(0)\le\int_0^1 (x+f(x))\,\mathrm{d}x=\frac32$. Now the right inequality implies $f(x)\le \frac52$ for each $x$. (Thanks to Nate Eldredge, who pointed in his comment, that this was missing in my original answer.)

To show that it is closed in sup-norm, you only have to show that if $f_n$ converges to $f$ uniformly and $f_n\in\mathcal F$, then the limit is in $\mathcal F$. We know that integral behaves well w.r.t. uniform convergence, see this questions. Proof of the fact that the condition $(\forall x,y\in [0,1])|f(x)-f(y)|\le |x-y|$ is preserved by uniform convergence is more-or-less standard. (In fact, in this part we only use pointwise convergence.)