Given two random variables $X$ and $Y$ that are $U[0,1]$, the CDF of their product $U$, where $U=XY$ is:
\begin{equation} \int_u^1 \bigg[\frac 1 x \times1\bigg] \, dx \quad \quad 0 < u < 1 \end{equation}
... and that's where I got stuck, why is the lower part of the integral $u$ and the term inside the integral $\frac{1}{x}$?
Yes, that is not quite right. The derivation of the probability density function (pdf) for $U$ goes:$$\begin{split}f_{XY}(u) &= \int_\Bbb R f_X(x)~f_Y(u/x)~\left\lvert\dfrac{\partial (u/x)}{\partial u}\right\rvert\mathsf d x\quad\big[u\in[0;1]\big]\\ &= \int_\Bbb R\mathbf 1_{0\leqslant x\leqslant 1}\mathbf 1_{0\leqslant u/x \leqslant 1}\lvert\tfrac 1x\rvert\mathrm d x\\ &= \int_{u\leqslant x\leqslant 1}\dfrac1x\mathsf d x\cdot\mathbf 1_{0\leqslant u\leqslant 1}\\ &= \int_{u}^{1}\dfrac1x\mathsf d x\cdot\mathbf 1_{0\leqslant u\leqslant 1}\\ & = -\ln (u)\mathbf 1_{0< u\leqslant 1}\end{split}$$
Then the CDF is found by $$\begin{split}F_{XY}(u)&=\int_0^u f_{XY}(x)\mathsf d x\\ & = \int_0^u-\ln(x)\mathsf d x\cdot\mathbf 1_{0< u\leqslant 1}+\mathbf 1_{1<u}\\ &= u(1-\ln (u))\cdot\mathbf 1_{0< u\leqslant 1}+\mathbf 1_{1<u}\end{split}$$