Show that $\int_{\mathbb{R}} F(x) F(dx)=\frac{1}{2}$ if $F$ is a continuous distribution function

You can use this result to show the following $$ I = \int_\Bbb R F(x)\mathrm dF(x) = F^2(x)|^\infty_{-\infty} - \int_\Bbb R F(x)\mathrm dF(x) = 1-I\implies I = \frac12 $$

Since Xiao provided another solution, I'll think it's worth telling you about a third one which is a neat probabilistic one-liner. Since $X$ is continuous, $F(X)\sim U[0,1]$ and hence your integral is an expectation of a uniformly distributed random variable on $[0,1]$ which is of course $\frac12$.

Here is the solution that I think you are looking for, recall from your other problem $$\int_\Omega f(x) dP = \int_0^\infty P\big(\{f>t\}\big) dt,$$ observe that $\mu_F$ is a probability measure on $\mathbb{R}$, so let $\Omega := \mathbb{R}$ and $P:= \mu_F$, we get $$\int_{-\infty}^\infty F(x) d\mu_F = \int_0^\infty \mu_F\big(\{F>t\}\big) dt =\int_0^1 \mu_F\big(\{F>t\}\big) dt,$$ because for $t>1$, the set $\{F>t\}$ is an empty set, thus $\mu_F\big(\{F>t\}\big)=0$ .

And for $t\in [0,1]$, I claim that $\mu_F\big(\{F>t\}\big) = 1-t$.

Here notice that since $F$ is a cdf, thus it is monotone, then the set $\{F>t\}$ is an interval $(r, \infty)$ in $\mathbb{R}$. (When $F$ is strictly increasing, then $F$ is invertable and we can take $r = F^{-1}(t)$). From the definition of L-S measure on intervals and $F$ is continuous, we have $$\mu_F((r, \infty)) = F(\infty) - F(r) = 1-t$$ then $$\int_{-\infty}^\infty F(x) d\mu_F =\int_0^1 1-t\;dt = 1/2.$$