Embedding $S_n$ into $A_{n+2}$

You are correct. You can write your map as $\phi :S_n \to S_n \times S_2 \subset S_{n+2}$ by $\phi(\sigma) = (\sigma, \tau^{|\sigma|})$, where $\tau$ is the nontrivial permutation in $S_2$. Then $\phi$ is obviously injective and

$$\phi(\sigma \rho) = (\sigma \rho, \tau^{|\sigma\rho|}) = (\sigma, \tau^{|\sigma|})\cdot (\rho, \tau^{|\rho|}) = \phi(\sigma)\phi(\rho)$$

as $|\sigma \rho| = |\sigma|+ |\rho|$. As $\phi(\sigma)$ is an even permutation for all $\sigma$, we have $\phi :S_n \to A_{n+2}$.

You know how to put $A_n$ inside $A_{n+2}$, i.e. produce an injective homomorphism $A_n\hookrightarrow A_{n+2}$, simply send $\sigma\in A_n$ to $\hat \sigma i=\sigma i $ if $1\leqslant i\leqslant n$ and $\hat\sigma i=i$ else. It remains that you fit in $C_n=S_n\setminus A_n$. But if $C_n$ is a copy of $B_n$ inside $S_n$, then $(n+1,n+2)B_n\subset A_{n+2}$ consists of even permutations. You can see $ A_n\cup (n+1,n+2)C_n$ is a subgroup of $A_{n+2}$ -- note that $(n+1,n+2)$ doesn't interact with anything in $C_n$ or $A_n$; and $C_n\cup A_n=S_n$ -- $S_n\simeq A_n\cup (n+1,n+2)C_n$ so that $S_n\hookrightarrow A_{n+2}$.