Compositions of homotopic maps are homotopic I'm reading some lecture notes on homotopy, and the author has just proved the theorem:   


  If $f_{0} \simeq f_{1}$ and $g_{0} \simeq g_{1}$, then $g_{0} \circ f_{0} \simeq g_{1} \circ f_{1}$


Where, for each $i \in \{0,1\}$, $f_{i} : X \to Y$ and $g_{i}: Y \to Z$, and $\simeq$ denotes homotopy equivalence.  

He does this by showing $g_{0} \circ f_{0} \simeq g_{0} \circ f_{1} \simeq g_{1} \circ f_{1}$
For the first equivalence he uses the homotopy $g_{0} \circ F$ (where $F$ is a homotopy from $f_{0}$ to $f_{1}$) and for the second he uses $G \circ (f_{1} \times Id_{[0,1]})$. I completely understand that. But I don't see why you couldn't (or wouldn't) use the quicker:

$$H(x,t)=G(F(x,t),t)$$

This is continuous (unless there's something key I've misunderstood), and $H(x,0)=G(F(x,0),0)=g_{0}(F(x,0))=g_{0}(f_{0}(x))$ and similarly for $t=1$.
There's no strong reason not to use $G(F(x,t),t)$.  One reason the author may not have is that they may have had in mind the general principle that you can concatenate homotopies, and that this often lets you build homotopies that "do multiple things" by doing each one separately and concatenating the pieces along the time coordinate.  In this particular example, it is actually possible to do both of them at once as you have pointed out, but it isn't always (see for instance my answer to this question).
Edit: as in the comments, this is clearly false.

I think the difficulty lies in showing rigorously that $G(F(x,t),t)$ is continuous.

While I see no reason that it shouldn't be, I can't see a way to show it is continuous that is as simple as the appeal to the glueing lemma that shows the construction given in the notes is continuous.