Show there is a path in X $\times$ Y if and only if there is a path in X and a path in Y (a) Let $(x_1,x_2) \in X$ and $(y_1,y_2) \in Y$. Show that there is a path from $(x_1,y_1)$ to $(x_2, y_2)$ if and only if there is a path $x_1$ to $x_2$ in $X$ and a path $y_1$ to $y_2$ in $Y$.
(b) show that $X \times Y$ is path connected if and only if $X$ and $Y$ are both path connected.
Firstly, I'm not sure about the difference between questions and (a) and (b). I would have assumed that proving (a) would also prove (b).
I went about this question as follows:
First I prove that if there is a path $x_1$ to $x_2$ in $X$ and a path $y_1$ to $y_2$ in $Y$ then there is a path $(x_1,y_1)$ to $(x_2, y_2)$ in $X \times Y$.
Suppose $\exists$ a continuous function $g: [0,1] \to X$ and a continuous function $h: [0,1] \to Y$. Then $X$ and $Y$ are path connected. Suppose $x_1$ and $x_2$ $\in U \subset X$ and $y_1$ and $y_2$ $\in V \subset Y$. Then $$(x_1,y_1), (x_2,y_2) \in U \times V \subset X \times Y$$ hence there exists a continuous function $f: [0,1] \to X \times Y$
For proving the converse, could I just say that there exists a projection map $\pi_1(x_1,y_1) = x_1$ and $\pi_1(x_2,y_2) = x_2$ which creates a continuous map from $[0,1] \to X$? And the same for $Y$?
 A: Path-connectedness is much more than just saying that there exists a continuous function $g:[0,1]\to X$, especially considering there is always such a map when $X$ is non-empty (namely, take a constant map).  Path-connectedness says that for every two points $x$ and $y$ in $X$, then there exists a continuous map $f:[0,1]\rightarrow X$ such that $f(0)=x$ and $f(1)=y$.  
Really the entire question boils down to the fact that $f:Z\to X\times Y$ is continuous if and only if there exists $g:Z\rightarrow X$ and $h:Z\rightarrow Y$ with $g,h$ continuous and $f(z)=(g(z),h(z))$ for all $z\in Z$.  What you say with regards to the projections (which you will recall, are continuous maps) helps to prove this (if you haven't done so already).  Then apply this to paths in $X$, $Y$, and $X\times Y$.
To prove the direction that if $g,h$ are continuous, we can actually start by proving that if we have maps $\varphi: A\rightarrow X$ and $\psi: B\rightarrow Y$, then the map $\varphi\times \psi: A\times B \rightarrow X\times Y$ defined by $\varphi\times \psi: (a,b)\mapsto (\psi(a),\psi(b))$ is continuous; this follows by noting that for $U\subset X$ and $V\subset Y$, then $(\varphi\times\psi)^{-1}[U\times V]=\varphi^{-1}[U]\times \psi^{-1}[V]$. Next, show that the map $\chi: Z\rightarrow Z\times Z$ defined by $\chi(z)=(z,z)$ is continuous (show $\chi^{-1}[U\times V]=U\cap V$).  Then taking $A=Z=B$, we'd see that $f=(\varphi\times \psi)\circ \chi$ is the composition of continuous functions, and hence continuous itself.
