There is only one natural way to construct a map $X \times Y \rightarrow X$ for topological spaces. Consider the functors $\text{id}_{\textbf{Top}}$ and $(-) \times Y: \textbf{Top} \rightarrow \textbf{Top}$, can we show that there is only one natural transformation $\pi: (-)\times Y \implies \text{id}_{\textbf{Top}}$? Namely the one given by $\pi_X,$ the projection $X \times Y \rightarrow X$. 
I intuitively suspect this to be the case but why is this true for $\textbf{Top}$ and not for $\textbf{Ab}$ for example? In $\textbf{Ab}$ we have zero morphisms so two nonequal natural transformations $\pi$ and $0: (-)\times A \implies \text{id}_{\textbf{Ab}}$, so the terminal object not being initial in $\textbf{Top}$ i.e zero objects not existing has to play some part in showing the uniqueness of $\pi$.
 A: Indeed, the projection is the only natural transformation $f_X:X\times Y\to X$, namely the projection. It's not hard to see, and as you suspected the proof involves the terminal element.
For any $x\in X$, we have the map $\varphi_x:\ast\to X$ sending the point to $x$. Then the commutative diagram
$$\require{AMScd}
\begin{CD}
\ast\times Y @>{f_\ast}>> \ast\\
@V{\varphi_x\times id_Y}VV @VV{\varphi_x}V \\
X\times Y @>{f_X}>> X
\end{CD}$$
shows that $f_X(x,y)=x$ for all $y\in Y$.
As I mentioned in my comment, this is related to the matter of natural transformations of the identity functor. In this case the key point is that the terminal object not only is non-zero, but actually represents the forgetful functor.
A: The terminal object is not initial, in fact the terminal object represents the forgetful functor, and this will be useful. 
Indeed if $\theta$ is such a transformation, $\theta_* : *\times Y\to *$ is clearly determined, and if $X$ is any space, and $x\in X$, then the map $*\to X$ sending the point to $x$ induces $*\times Y\to X\times Y$ sending $(*,y)\mapsto (x,y)$. So the naturality of $\theta$ applied to $*\overset{x}\to X$ will give you what you want. 
