I mostly just want to check my work.
Claim: If $h,h': X \rightarrow Y$ are homotopic and $k,k':Y \rightarrow Z$ are homotopic, then $k \circ h$ and $k' \circ h'$ are homotopic.
Proof: Let $H(t,s): I \times X \rightarrow Y$ be a homotopy from $h$ to $h'$, and let $K(t,s): I \times Y \rightarrow Z$ be a homotopy from $k$ to $k'$. Define $G = K(t,H(t,s))$. If $U$ is open in $Z$, $K^{-1}(U) \cap$ dom $K$ is open in $I \times Y$ as $K$ is continuous. It follows that $K^{-1}(U) \cap$ dom $K$ is a union of products of open sets in $I$ and $Y$ respectively. If $V$ is open in $Y$, then $H^{-1}(V) \cap$ dom $V$ is open in $I \times X$. Thus open sets pull back to open sets via $G$. Thus $G$ is continous. Lastly, $G(0,s) = K(0,H(0,s)) = K(0,h(s)) = k(h(s))= (k \circ h)(s)$ and $G(1,s) = K(1,H(1,s)) = K(1,h'(s)) = k'(h'(s)) = (k' \circ h')(s)$.
Is this correct? If so, why not? Or would you kindly direct me towards a more correct argument.
The argument is correct, very nice. You shorten your proof, by the way, by some by observing just that $G$ is continuous simply because it is a composition of continuous functions.