If $X$ has CDF $F$, find the CDF of $Y=X^2$ 
Let $X$ be a random variable with cumulative distribution function $F$. Let $Y=X^2$. Find the cumulative distribution function $F_Y$ of  $Y$ in terms of $F$.

First we observe that $0\leq X^2$, and hence $0\leq Y$. So if $t<0$ then
$$F_Y(t)=P(X^2\leq t)=0.$$
Now if $0\leq t$, then
$$F_Y(t)=P(X^2\leq t)=P(X\leq\sqrt t)=F(\sqrt t).$$
Therefore,
$$F_Y(t)=\begin{cases}0&\text{if }t<0\\F(\sqrt t)&\text{if }t\geq 0.\end{cases}$$
Notice that $F_Y$ is right-continuous, by construction.

Do you agree with my work above? Thank you for your time and appreciate any feedback.

What you have done is not correct. You are assuming that $X \geq 0$ which is not given.
Let $t \geq 0$. Note that for  $X^{2} \leq t$ iff $-\sqrt t \leq X \leq \sqrt t$. So $P(X^{2} \leq t)=P(X \leq \sqrt t)-P(X<-\sqrt t)$. This can be written as $F(\sqrt t)-F((-\sqrt t)-)$ where $F(x-)=sup_{y<x} F(y)$, the left hand limit of $F$ at $x$. . 
