Q:

$X \times Y$ path connected implies $X$ and $Y$ path connected.

$$\textbf{PROBLEM}$$
Suppose $X$ and $Y$ are topological spaces. If $X \times Y$ is path
  connected, then $X$ and $Y$ are path connected.

$$\textbf{ATTEMPT}$$
IF $X \times Y$ is path connected, then we can take a path $\gamma$ from $[0,1]$ to $X \times Y$ such that $\gamma(0) = (x_0, y_0)$ and $\gamma(1) = (x_1,y_1)$. If we define $\gamma_0(0) = (x_0,0)$ and $\gamma(1) = (x_1,0)$, then we have a path from $x_0$ to $x_1$ in $X$. Hence, $X$ is path connected. Similarly, $Y$ is path connected.
Another approach
We know $$\pi_X : X \times Y \to X$$
$$ \pi_Y : X \times Y \to Y$$
are homeomorphisms and since path-connectedness is topological invariant, then we get out conclusion.
Are they correct? Can I get some feedback? thanks

A:

Your first version doesn’t really make sense: you have the path $\gamma$ such that $\gamma(0)=\langle x_0,y_0\rangle$, so you can’t redefine $\gamma(0)$ to be $\langle x_0,0\rangle$. You could try to define a new path, but in that case you have to give it a different name. Moreover, if $\gamma(0)=\langle x_0,0\rangle$, you’re not constructing a path in $X$, but rather in $X\times\Bbb R$ or something similar.
Your second approach is a much better idea, except that the projection maps are not in general homeomorphisms. They are, however, continuous, and you know from the earlier question that continuous maps preserve path-connectedness.

A:

Your second argument is the more elegant because it contains your first argument. All you need to finish it off is the fact that the image of a path-connected set under a continuous map is path-connected and that the projection maps are continuous surjections (although not homeomorphisms).

