Let $r$ and $s$ be the generator $Q_8$ described in Section $5$. Let $\phi$ be the map from $Q_8$ to $\operatorname{GL}_2(\mathbb{C})$ defined on generators by $$ \phi(r) = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} \text{ and } \phi(s) = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} $$ which extends to a homomorphism. Prove that $\phi$ is injective.
Typically I would let $a,b\in Q_8$, such that $\phi(a)=\phi(b)$, and then show that $a=b$. However, I was reading only that you can prove injectivity by showing the $\ker\phi=e_{\operatorname{GL}_2(\mathbb{C})}=I$. Now I don't typically use the kernel so can someone give me some pointers on how to go about this question and future ones of this kind?
What you have read is correct as a general approach, but because $Q_8$ is such a small group, it's better to simply list elements of the form $\phi(q)$ for $q \in Q_8$.
In fact, I recommend the following: compute $$ \phi(r^2)= \phi(r)^2 = \pmatrix{-1&0\\0&-1}, \\ \phi(r^3) = \phi(r^2) \cdot \phi(r) = \pmatrix{i & 0\\0&-i},\\ \phi(s^3) = \phi(s^2) \cdot \phi(s) = \phi(r^2) \cdot \phi(s) = \pmatrix{0&1\\-1&0},\\ \phi(rs) = \phi(r) \cdot \phi(s) = \pmatrix{0&-i\\i & 0},\\ \phi((rs)^3) = \phi((rs)^2) \cdot \phi(rs) = \phi(r^2) \cdot \phi(rs) = \pmatrix{0&i\\-i&0}. $$ Because $\phi$ is a homomorphism, $\phi(1)$ must be the identity matrix. With this, you will have shown that $\phi$ maps the elements of $Q_8$ to distinct elements and is therefore indeed injective.
Proof that $(rs)^2 = r^2$: note that $$ (rs)^2 = (rsr)s = ss = s^2 = r^2. $$