Symmetry of "is homotopic to" detail in the proof 
Let $f,g:X\rightarrow Y$. If $f$ is homotopic to $g$ then $g$ is homotopic to $f$.

Let $F:X\times I\rightarrow Y$ be a homotopy from $f$ to $g$ so $F(x,0)=f(x)$ and $F(x,1)=g(x)$ for all $x \in X$ and $F$ is continuous. Define $H:X\times I\rightarrow Y$ by $H(x,t)=F(x,1-t)$. We see $H(x,0)=g(x)$ and $H(x,1)=f(x)$. I am wondering what can I use to justify $H$ being continuous. I see it looks like $H$ in a way is a composition of functions, like we are doing $t\mapsto 1-t$ to get from $F(x,t)$ to $H(x,t)=F(x,1-t)$. I was hoping to just prove something in general like:

If $F:X\times Y\rightarrow Z, f:X\rightarrow X$ and $g:Z\rightarrow Z$ are each continuous functions where $f$ and $g$ are bijections, then $H:X\times Y\rightarrow Z$ by $H(x,y)=F(f(x),g(y))$ is continuous.

I think this may be an uglier form of something that I could have seen before with continuous functions.
Concerning the original $H$ function, I tried to show it was continuous by picking some $U$ open in $Y$ and trying to find out what the set $H^{-1}(U)$ would be. I am not sure how to use that $t\mapsto 1-t$ is a continuous bijection on $I=[0,1]$ and that $F$ is continuous to write $H^{-1}(U)$ as an open set in $X\times I$.
 A: Here is the way I would think about it. To define a map $Z \to X \times Y$, I need to give maps $Z \to X$ and $Z \to Y$. So, I claim that the map $X \times I \to X \times I$ defined by $f(x,t)=(x,1-t)$ is continuous. This is because the map $X \times I \to I \to I$, where the first arrow is projection and the second arrow is the map $t \mapsto 1-t$, is continuous (because the composition of continuous functions is continuous!).
So, you can view your backwards homotopy at the composition $X\times I \to X \times I \to Y$, where the first arrow is the map  defined above and the second is your homotopy you already have. The moral is that once you prove that one thing is continuos, closely related things should be continuous to and this should follow formally. 
A: I think you're missing a much more general statement: the composition of any two continuous functions is continuous.  To avoid repurposing your letters, consider the functions $a: M \to N$ and $b: N \to P$, and suppose that both $a$ and $b$ are continuous.  Let $V \subseteq P$ be any open set.  Since $b$ is continuous, $b^{-1}(V)$ is open in $N$; since $a$ is continuous, $a^{-1}(b^{-1}(V))$ is open in $M$.  Finally, it is immediate to check that $(b\circ a)^{-1}(V) = a^{-1}(b^{-1}(V))$ for any functions between sets.
It is not difficult to check directly that the endofunction $(x,t) \mapsto (x,1-t)$ of $X\times [0,1]$ is continuous.  A more general statement is that if $a: M \to N$ and $b : P \to Q$ are both continuous, then $a\times b : M \times  N \to P \times Q$ is continuous.  This follows from the following characterization of the product topology: a subset $V \subseteq P \times Q$ is open iff it is a (possibly infinite) union of "open rectangles" $V_P \times V_Q$, where $V_P \subseteq P$ and $V_Q \subseteq Q$ are both open.  Then use the fact from set theory that preimages of (even infinite) unions are unions.
A: Thank you James and Theo for your help. I will type up here what I got from looking at both your answers.
I take Theo's claim:

$a:M\rightarrow N$ and $b:P\rightarrow Q$ are continuous functions implies $g=a\times b:M\times P\rightarrow N\times Q$ by $g(m,p)=(a(m),b(p))$ is continuous.

I see this is a continuous function.
From here as $g_1:X\rightarrow X$ by $g(x)=x$ and $g_2:I\rightarrow I$ by $g_2(x)=1-t$ are both continuous functions, so by above we have then $g:X\times I\rightarrow X\times I$ by $g(x,t)=(x,1-t)$ is a continuous function.
For the problem I originally posted:
Since we know $g:X\times I\rightarrow X\times I$ via $(x,t)\mapsto (x,1-t)$ is continuous and we know the function $F:X\times I\rightarrow Y$ is continuous, it follows $H=F\circ g$ is a continuous function as I wanted to show.
