A subset $A \in \mathbb{R}^n$ is convex if $\forall p,q \in A$, $\{{tp+(1-t)q | 0 \le t \le 1}\} \subset A$.
I have two paths $\alpha, \beta : [0,1] \rightarrow A$, where $\alpha(0)=\beta(0)=p$ and $\alpha(1)=\beta(1)=q$, so that $\alpha$ and $\beta$ are paths from $p$ to $q$ in $A$.
To show that these paths are homotopic, I believe I need to build a continuous map $H: [0,1] \times [0,1] \rightarrow A$ with $H(t,0)=\alpha(t) \,,\, H(t,1)=\beta(t)\,,\, H(0,s)=p\,,\, H(1,s)=q$.
So far, I have the following homotopy:
$H(t,s)=\{(1-s) \cdot \alpha(t)+s \cdot \beta(t) \,| \,0\le s\le 1, 0 \le t \le 1\} $
This allows $H(t,0)=\alpha(t)$ and $H(t,1)=\beta(t)$, but $H(0,s)=(1-s)\cdot p +s \cdot p$, which does not match with the convexity property.
I initially thought I could fix this with $H(t,s)=\{(1-s) \cdot \alpha(1-t)+s \cdot \beta(t) \,| \,0\le s\le 1, 0 \le t \le 1\} $, but then $H(t,0) \ne \alpha(t)$.
How can I get around this problem?
Your homotopy works. I'm not sure what you mean when you say $(1-s)\cdot p + s\cdot p$ does not match the convexity property - it sure looks like it does, it's just a bit of a degenerate case, since the two endpoints of the segment are the same. I assume that you consider it obvious that $H$ is, in fact, continuous and we only need to show that $H(t,s)$ remains in the set. To do this more generally, let $\alpha(t) = p_1$ and $\beta(t)=p_2$. Then, $H(t,s)=(1-s)p_1+sp_2$. Essentially, considering $t$ as fixed and $s$ as varying in $[0,1]$, this represents the line segment between $p_1$ and $p_2$ - which ought to stay within the set, by convexity. Moreover, if you just look at your property for convexity, replacing $p$ by $p_1$ and $q$ by $q_1$ and $t$ by $s$, that this line stay within the set is exactly the property given for convexity.