If $X \sim \text{U}(0,1)$, what is the distribution of $Y = \max(X,1/X)$? I know for this particular problem, $Y = \max(X,1/X) = 1/X$, whose distribution can be easily attained directly. However, I've been trying another approach by starting with this
$$P(Y \leq y) = P(X \leq y,\frac{1}{X} \leq y)=P(\frac{1}{X}\leq y| X \leq y)P(X \leq y),$$
and I've been stuck. I would be appreciate if someone could show me how it could be proceeded. Thank you very much.
As you have acknowledged in your question, this problem is trivial in the case where $0 < X < 1$. In the more general case (taking $\tfrac{1}{0} = +\infty$ for definiteness) you have:
$$Y = \max(X, \tfrac{1}{X}) = \begin{cases} \tfrac{1}{X} & & \text{if }X < -1 \\[6pt] X & & \text{if }-1 \leqslant X < 0 \\[6pt] +\infty & & \text{if } X = 0 \\[6pt] \tfrac{1}{X} & & \text{if }0 < X < 1 \\[6pt] X & & \text{if }X \geqslant 1 \\[6pt] \end{cases}$$
If you restrict this to the case where $X>0$ (which includes the last two categories), you have:
$$\begin{equation} \begin{aligned} F_y(y) \equiv \mathbb{P}(Y \leqslant y) &= \mathbb{P}(\max (X, \tfrac{1}{X}) \leqslant y) \\[6pt] &= \mathbb{P}(X \geqslant \tfrac{1}{y}, 0 < X < 1) + \mathbb{P}(X \leqslant y, X \geqslant 1) \\[6pt] &= \mathbb{P}(\tfrac{1}{y} \leqslant X < 1) + \mathbb{P}(1 \leqslant X \leqslant y). \\[6pt] \end{aligned} \end{equation}$$
And of course, in the case where $0<y<1$ this probability is zero, as expected.
