You can start by showing that if $f \colon [a,b] \rightarrow \mathbb{R}$ is a bounded function and $c \in (a,b)$ such that both $f|_{[a,c]}$ and $f|_{[c,b]}$ are Riemann integrable, then $f$ is Riemann integrable and
$$ \int_a^b f(x)\, dx = \int_a^c f(x)\, dx + \int_c^b f(x) \, dx. $$
Then, prove that for any $a < x_0 < b$ and $y_0 \in \mathbb{R}$, the function
$$ f_{x_0,y_0}(x) := \begin{cases} y_0 & x = x_0 \\ 0 & x \neq x_0 \end{cases} $$
is Riemann integrable on $[a,b]$ with integral zero. The argument will be the same argument as for your $f_1$.
Now, choose some $\frac{1}{2} < a_1 < 1$. On $[a_1,1]$, your function $f$ is just $1 - f_{1,1}(x)$ and thus, is Riemann integrable. On $[\frac{1}{2},a_1]$, your function is just $1 - f_{\frac{1}{2},1}$ and thus is also Riemann integrable. By the result above, $f|_{[\frac{1}{2},1]}$ is Riemann integrable. Continuing this way inductively, you can see that $f|_{[\frac{1}{n},1]}$ is Riemann integrable for all $n \in \mathbb{N}$. Alternatively, if you already know that a function with finitely many discontinuities is Riemann integrable, you can skip all of the above.
Finally, using the definition of the Riemann integrable directly, show that if $f \colon [a,b] \rightarrow \mathbb{R}$ is a bounded function such that $f|_{[a + \frac{1}{n},b]}$ is Riemann integrable for all sufficiently large $n$, then $f$ is Riemann integrable and
$$ \int_a^b f(x) \, dx = \lim_{n \to \infty} \int_{a + \frac{1}{n}}^b f(x)\, dx. $$
Naturally, there are only finitely many $n$ such that $\frac{1}{n} \geq \frac{\epsilon}{2(b-a)}$, because $\frac{1}{n}$ is a function that decreases with $n$. Now, since our domain is $(0,1]$, we must have $m \leq n$, which means only finitely many m exist, so only finitely many $x=\frac{m}{n}$ exist. Let the number of such $x$ be $N$.
Take any partition $P$ of $[0,1]$, $ P = x_0 < x_1 < ... < x_k=1$. Note that:
$$
S(f,P) = \sum (x_{i+1} -x_i) M_i, M_i=\displaystyle\sup_{[x_{i+1},x_i]} f(x)
$$
$$
s(f,P) = \sum (x_{i+1} -x_i) m_i, m_i=\displaystyle\inf_{[x_{i+1},x_i]} f(x)
$$
But then $m_i=0$ for all $i$, since in every interval there is some irrational number. So $s(f,P)=0$. Now note that $f(x) \leq 1/2$ except at $x=1$. In order to take care of that, we will consider a partition of [0,1] where $\delta > 2N$. Let it have say T partition points.Now, we have that:
$$
S(f,P) = \sum_{i=0}^{T-1} (x_{i+1} -x_i) M_i, M_i=\displaystyle\sup_{[x_{i+1},x_i]} f(x) \leq \sum_{i=0}^{T-2} (x_{i+1} -x_i) * 1/2 + (1-x_{T-1})
$$
Finish it from here.
Best Answer
Well, first figure out what you think the integral should be (just as you would figure out what you think a limit should be before you attempt an $\varepsilon$-$\delta$ proof), and call it $I$. Hint: the function takes the value $1$ pretty much everywhere.
Then, you wish to establish that $I$ actually is the integral. You need to show that $I(f) \le I \le U(f)$, where $I(f)$ is the supremal value of lower sums of $f$ over all partitions, and $U(f)$ is the infimal value of upper sums of $f$ over all partitions.
It's not difficult to show that $U(f) = I$. Just show $U(f, P) \ge I$ for all partitions $P$, and find a particular partition such that $U(f, P) = I$ (this shouldn't be hard: literally any partition $P$ should do!).
To establish that $L(f) = I$, you only need to show that $L(f) \ge I$, since we automatically have $L(f) \le U(f)$. You need to construct a sequence of partitions $P_n$ such that $L(f, P_n) \rightarrow I$. Since $x = 0$ is the one and only problematic point, I suggest isolating $x = 0$ in an increasingly small part, for example, $$P_n = \lbrace [0, 1/n], [1/n, 1] \rbrace$$ should do the trick! Compute $L(f, P_n)$. It should be a very simple sequence in terms only of $n$, and it should approach $I$.
Good luck!