Using gluing/pasting lemma to show two paths are homotopic? 
($\dot I = \{0,1\}$)
The homotopy I've constructed is:
 $$G(t_1, t_2) =
\begin{cases}
\alpha(t_1),  & \text{if $(t_1,t_2) \in I \times \{0\}$} \\
\gamma * \beta* \delta^{-1}(t_1) , & \text{otherwise}
\end{cases}$$
But this isn't correct since the domain of this function isn't a union of two open or closed sets as per the gluing lemma.
Anyone have any ideas as to how to re-write the domain so that this becomes the required homotopy?  
 A: You are correct that the gluing lemma doesn't work:  i.e. $G$ is not continuous unless $\gamma$ and $\delta$ are constant paths.  The question you ask in the bottom is 'can I do something differently to make this map continuous'.  I think the answer is no - the information given $F$ is not used in constructing the homotopy and this information is critical.
One thing related to convexity is that we have the map $r_t: I \times I \to I \times I$, $(s_1,s_2) \mapsto (s_1,ts_2)$. which is a deformation retraction of the inclusion $I \times \{0\} \hookrightarrow I \times I$.  We can use this evidently continuous retraction to construct your homotopy.
Let $\sqcap: I \to I \times I$ be the path with $F \circ (\sqcap)= \gamma * \beta * \delta^{-1}$.  Then because $r_t$ is a deformation retraction $\gamma * \beta * \delta^{-1}= F \circ r_0 \circ  (\sqcap) \cong^{homotopic} F \circ r_1 \circ (\sqcap)  \cong^{homotopic} \alpha $.
The homotopy of the first equivalence is  $h_t=F \circ r_t \circ (\sqcap)$.  The homotopy of the second equivalence is the homotopy (identity path at $(0,0)$) $* \alpha *$(identity path at $(1,0)$) $ \cong \alpha$.

Edit:  Above I wrote a crap answer.  Prof. Brown has the right idea.  Let $\text{_____}$ be the path given by $\text{_____}(t)=(t,0)$ in $I \times I$.  Then there is a homotopy $g_t$ from $\sqcap$ to $\text{_____}$.
$h_t=F \circ g_t$ is a homotopy from $\gamma * \beta * \delta^{-1}$ to $\alpha$.
A: Let $S$ be a simply connected space and let $g: S \to X$ be a map. Let $a,b: [0,1] \to S$ be paths in $S$ with the same end points. Then $ga,gb: [0,1] \to X$ are homotopic rel end points. 
Proof Since $S$ is simply connected there is a homotopy $H_t:a \simeq b$ rel end points. Then $gH_t:ga \simeq gb$. 
In the case we need, $S$ is a square. It is convex and we can define $H_t(s)= (1-t)a(s) + t b(t)$. Then $H_0(s) =a(s), H_1(s) = b(s)$. 
