I am asked to show that for $1 \leq x,y$ and an integer $q$, we have:
$S(x,x+y,q) = |\{x < n \leq x + y \mid n \text{ is comprime to } q\}| = \frac{\phi(q)y}{q} + O(2^{\omega(q)})$, where:
$\omega(q)$ is the number of distinct prime factors of $q$.
We may note that $S(x,x+y,q) = S(1,x+y, q) - S(1,x,q)$ and also:
$S(1,x,q) = \phi(q) \lfloor \frac{\lfloor x \rfloor}{q} \rfloor - 1 = \frac{\phi(q)x}{q} + O(\frac{\phi(q)}{q}) + O(\phi(q)) -1 = \frac{\phi(q)x}{q} + O(\phi(q)) - 1$
This is almost the required form and if I can show that: $\phi(q) = O(2^{\omega(q)}) $ then I'd be done.
However, I am not convinced that this is actually true, considering we may use Euler's Porduct Formula to see:
$$\phi(q) = q\prod_{\:\:\:p \mid q \\ p \text{ prime}}\left(1 - \frac{1}{p}\right)$$
And I believe that although we take the product over exactly $\omega(q)$ many terms, each term can easily be much larger than $2$.
This question is likely meant to be solved using Sieve methods, but I can't see anything in the notes that immediately tells me I might be able to shrink this error term. Is there any standard way I may be able to achieve this error term?
As you have essentially noticed, it suffices to consider intervals of the form $[1,x]$. Splitting every such interval into a system of intervals of length $q$ and a "residual" interval of length smaller than $q$, and observing that every interval of length $q$ contains exactly $\phi(q)$ integers co-prime with $q$, it remains to show that every interval $[1,R]$, with $1\le R<q$, contains $q^{-1}\phi(q)R+O(2^{\omega(q)})$ integers co-prime with $q$. This can be done using the Mobius function; namely, denoting by $\tau(q)$ the number of sqare-free divisors of $q$, \begin{align*} |\{n\in[1,R]\colon (n,q)=1\}| &= \sum_{n=1}^R \sum_{d\mid (n,q)} \mu(d) \\ &= \sum_{d\mid q} \mu(d) \left\lceil \frac Rd\right\rceil \\ &= R \sum_{d\mid q} \frac{\mu(d)}d + O(\tau(q)) \\ &= \frac{\phi(q)}q\,R + O(2^{\omega(q)}). \end{align*}