Let $P=(0,1)$ and $Q=(4,1)$ be points on the plane. let $A$ be a point which moves on the $x$-axis between the point $(0,0)$ and $(4,0)$. let $B$ be a point which moves on the line $y=2$ between the points $(0,2)$ and $(4,2)$. Consider all the possible paths consisting of the line segments $\overline{PA}, \overline{AB}, \overline{BQ}$. What is the shortest possible length of such a path?

My guess/method would be to let $P'=(0,-1)$ and $Q'=(4,3)$ and draw a straight line from $P'$ to $Q'$. Wherever that line passes the x axis and the line $y=2$ would be your points for $A$ and $B$.
And the length of the path would be $4\sqrt{2}$
suppose the required minimum length path is $PABQ.$ let $Q_ = (0, -1), Q+ = (4,3)$ the mirror images of $Q(4,1), P(0,1)$ on the mirrors $y = 2, y = 0$ respectively.
we claim that $Q_ABQ+$ is a straight line and the length $4\sqrt 2$ of $Q_Q+$ is the shortest path.
proof of the claim: look at the triangle $Q_AB.$ by triangle inequality, $Q_B \le Q_A + AB.$ equality occurs when $Q_AB$ is a line. in the same way, by looking at the triangle $ABQ+,$ you can show $ABQ+$ is a line. that proves the claim.