Depends on what you're using as your definition of $D_8$.
One way would be to show that if you take $V=C_2\times C_2$, $\langle \alpha \rangle$ injects into the automorphism group of $V$, which is isomorphic to $GL_2(\mathbb{F}_2)$, by mapping $\alpha$ to $$\left(\begin{array}{cc}1&1\\0&1\end{array}\right).$$
Then show that the semidirect product defined by this automorphism is isomorphic to $D_8$ by whatever presentation you're using.
You could try the same idea using permutation group notation, taking $V=\langle(1,2),(3,4)\rangle$ and mapping $\alpha$ to the automorphism $$\varphi_\alpha:\left\{\begin{array}{l}(1,2)\mapsto (1,2)(3,4) \\ (3,4)\mapsto (3,4)\end{array}\right. .$$
Or yet another variation would be the presentation $$\langle r,s,t|r^2=s^2=t^2=[r,s]=[s,t]=1,r^t=rs\rangle.$$
It all depends how you want to do it, but this automorphism is the key.
More generally:
Theorem. If $G$ has a subgroup of index two, $H$ say, and there exists an element $g\not\in H$ of order two then $G$ splits as a semidirect product $G=H\rtimes\mathbb{Z}_2$.
This is because $\langle g\rangle\cap H=1$, and the other (internal) semidirect product conditions follow because $H$ has index two (so is normal, and so on).
So, in order to answer your question you simply need to find an element of order two in $S_n$ which is not contained in $A_n$.
Best Answer
$\mathbb{Z}\rtimes\mathbb{Z}_2$ is generated by an element of infinite order $a$ and an element of order two $b$ satisfying the relation $bab^{-1}=a^{-1}$. Try to find two elements of $\langle x,y\mid x^2,y^2\rangle$ with the same properties which also generate the whole group. Then you can establish an isomorphism.