I'm trying to prove the following statement: Let $m \geq 3, 2r \leq m+1, \{x_1,...,x_r\},\{y_1,...,y_r\}\subseteq \{1,2,...,m\}, 1\leq s\leq m$ then $\exists \sigma,\rho \in A_m$ with $\sigma x_r = \rho y_r = s$ and $\sigma\{x_1,...,x_{r-1}\}\cap\rho\{y_1,...,y_{r-1}\}=\emptyset$.
To construct such permutations take $\sigma'=(x_r s)\Pi_{i=1}^{r-1}(x_i 2i)$ and $\rho'=(y_r s)\Pi_{i=1}^{r-1}(y_i 2(i-1)+1)$ (for simplicities sake we dont worry about the case $s=2j$ or $s=2(j-1)+1$ for $j\leq r-1$ as this collision is easily avoided by skipping to the next odd/ even number). Now $sgn(\sigma')=sgn(\rho')=-1^r$ and if $r$ is odd we can just tack on a transposition that doesn't involve any of the $x_j$/$y_j$ respectively. Am I missing something here?
Your proof is basically fine, but it could use some punctuation and better grammar.
The "we can just tack on a transposition" part is where you end up using the assumption $m \ge 3, 2r \le m+1$ because it means there are definitely at least 2 items NOT contained in $\{x_1, \cdots, x_r\}$ for example.
I would have just assumed WOLOG $s = m$ and chosen $\hat \sigma(x_j) = j$ and $\hat \rho(y_j) = r-1+j$ for $1 \le j \le r-1$. This gives permutations $\hat \sigma, \hat \rho \in S_m$ which satisfy all the conditions except they may not be even. Then just tack on a transposition at the end if needed like you said. Basically the difference is that I wouldn't have bothered fiddling with odd vs even numbers for $\sigma(x_j)$ vs $\rho(y_j)$. Your way is still correct though.