Observe that
$$\left(-\infty, \dfrac{1}{x}\right) \cup \left\{\dfrac{1}{x} \right\}=\left(-\infty, \dfrac{1}{x} \right]$$
and that the sets $\left(-\infty, \dfrac{1}{x}\right)$, $\left\{\dfrac{1}{x} \right\}$ are disjoint.
Therefore, on the left-hand side,
$$\mathbb{P}\left[\left(-\infty, \dfrac{1}{x}\right) \cup \left\{\dfrac{1}{x} \right\}\right]=\mathbb{P}\left[\left(-\infty, \dfrac{1}{x}\right) \right] + \mathbb{P}\left(\left\{\dfrac{1}{x} \right\}\right) = \mathbb{P}\left(X < \dfrac{1}{x}\right)+0$$
because
$$\mathbb{P}\left(\left\{\dfrac{1}{x} \right\}\right) = 0$$
due to $X$ being a continuous random variable, and on the right-hand side, we simply have
$$\mathbb{P}\left[\left(-\infty, \dfrac{1}{x} \right] \right] = \mathbb{P}\left(X \leq \dfrac{1}{x}\right)$$
Hence, equating the two sides,
$$\mathbb{P}\left(X \leq \dfrac{1}{x}\right) = \mathbb{P}\left(X < \dfrac{1}{x}\right)$$
Convolution of two distributions.
$t_{x_0} = 0$
$t_{x_1} = 1$
$t_{h_0} = 0$
$t_{h_1} = 1$
Thus $$f_Y(t) = 0, t \le t_{x_0}+t_{h_0} ,$$
$$f_Y(t) = \int_{max(t_{h_0}, t-t_{x_1})}^{min(t_{h_1}, t-t_{x_0})} f_X(\tau)f_H(t-\tau)d\tau, \text{ } t_{x_0}+t_{h_0} \le t \le t_{x_1}+t_{h_1},$$
$$f_Y(t) = 0, t \ge t_{x_1}+t_{h_1} ,$$
THese translate to the following solution
First convolve two uniform distributions
$X(t) ~ U(0,1)$ and $H(t) ~ U(0,1)$
$$Y(t) = x(t).h(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau)d\tau$$
The above convolution reduces to
$y(t) = 0, t\lt 0$
$y(t) = \int_{max(0,t-1)}^{min(1,t)} d\tau , 0\lt t \lt 2$,
$y(t) = 0 , t \gt 2$
The middle one will have to split into two intervals, namely $0\lt t \lt 1$ and $1\lt t \lt 2$
$y(t) = 0, t\lt 0$
$y(t) = \int_{0}^{t} d\tau =t, 0\lt t\lt 1$,
$y(t) = \int_{t-1}^{1} d\tau = 2-t, 1\lt t\lt 2$,
$y(t) = 0 , t \gt 2$
Now $W(t) = Y(t). S(t)$ where $S(t) ~U(0,1)$
For $0\lt t\lt 1$, the bounds are
$t_{s_0} = 0$
$t_{s_1} = 1$
$t_{y_0} = 0$
$t_{y_1} = 1$
$$W(t) = \int_{max(0,t-1)}^{min(1,t)} \tau d\tau = \int_{0}^{t}\tau d\tau = \frac{t^2}{2}, 0\lt t\lt 1$$,
For $1\lt t\lt 2$, $S(t)$ convolves with $Y(t)$ on two intervals namely $(0,1)$ and $(1,2)$. For the interval $(0,1)$ the bounds are
$t_{s_0} = 0$
$t_{s_1} = 1$
$t_{y_0} = 0$
$t_{y_1} = 1$
and for the interval $(1,2)$ the bounds are
$t_{s_0} = 0$
$t_{s_1} = 1$
$t_{y_0} = 1$
$t_{y_1} = 2$
Thus $$W(t) = \int_{max(0,t-1)}^{min(1,t)} \tau d\tau + \int_{max(1,t-1)}^{min(2,t)} (2-\tau) d\tau$$ $$ = \int_{t-1}^{1}\tau d\tau + \int_{1}^{t}(2-\tau) d\tau$$ $$ = -\frac{1}{2}(2t^2-6t+3), 1\lt t\lt 2$$,
For $2\lt t\lt 3$, $S(t)$ convolves with $Y(t)$ on $(1,2)$. For the interval $(1,2)$ the bounds are
$t_{s_0} = 0$
$t_{s_1} = 1$
$t_{y_0} = 1$
$t_{y_1} = 2$
Thus $$W(t) = \int_{max(1,t-1)}^{min(2,t)} (2-\tau) d\tau $$ $$ = \int_{t-1}^{2} (2- \tau) d\tau$$ $$ = \frac{(t-3)^2}{2}, 2\lt t \lt 3$$
and finally $W(t) = 0, t\gt 3$
Thus the $W(t)$ is defined by
$W(t) = 0 , t\lt 0$
$W(t) = \frac{t^2}{2}, 0\lt t \lt 1$
$W(t) = -t^2+3t-\frac{3}{2}, 1\lt t \lt 2$
$W(t) = \frac{(t-3)^2}{2}, 2\lt t \lt 3$
$W(t) = 0 , t\gt 3$
Best Answer
You can find an expression for $F_{\xi^2}(x)$ now.
Evidently $F_{\xi^2}(x)=0$ for $x\leq 0$, and for $x>0$ we have:$$F_{\xi^2}(x)=P(\xi^2\leq x)=P(-\sqrt x\leq\xi\leq\sqrt x)=F_{\xi}(\sqrt x)-F_{\xi}(-\sqrt x)$$
If next to the CDF you also want a PDF then find it as derivative of the CDF.