I doubt that there is a trigonometry-free proof of the equidistribution
property (see statement below). Here, I will deduce the result in the OP from this
equidistribution property (because the solution proposed in the linked answer
is wrong, as has been pointed out).
Equidistribution property Let $I=[b,c]$ be a subinterval of
$[0,1]$. Then the frequency $f_n(I)=\frac{w_n(I)}{n}$ where
$w_n(I)$ is the number of $k\in[n]=\lbrace 1,2,3, \ldots, n\rbrace$
such that $\lbrace k\alpha \rbrace \in I$ satisfies
$f_n(I) \to {\sf length}(I)=c-b$ when $n\to +\infty$.
Solving your exercise with the equidistribution property
Let $\varepsilon > 0$. Let $\eta$ be another positive number, depending
on $\varepsilon$, to be defined later. Since $|f|$ is continuous on the
compact set $[0,1]$, it reaches a maximum there, which we will
denote by $M$. Since $f$ is continuous on the
compact set $[0,1]$, it is also uniformly continuous there. So there is
a $\delta >0$ such that for any $x,y\in [0,1]$
$$
|x-y| \leq \delta \Rightarrow |f(x)-f(y)| \leq \eta \tag{1}
$$
Let $q$ be an integer such that $\frac{1}{2^q} \leq \delta$. Then
the intervals $I_j=[\frac{j-1}{2^q},\frac{j}{2^q}] (1 \leq j \leq 2^q)$
make a subdivision of $[0,1]$. By the
equidistribution property, for each $j$ the sequence
$(f_n(I_j))_{n\geq 1}$ tends to $\frac{1}{2^q}$
when $n\to +\infty$. So there is an integer $n_0(q,\eta)$ such that
for any $n \geq n_0(q,\eta)$ and $1\leq j \leq 2^q$,
$$
\bigg|f_n(I_j)-\frac{1}{2^q}\bigg| \leq \eta \tag{2}
$$
Next, define
$$
X_{n,j}=\bigg\lbrace k\in [n] \ \bigg| \ \lbrace ka \rbrace \in I_j \bigg\rbrace \tag{3}
$$
If we put $d_n=\frac{\sum_{k=1}^{n} f(\lbrace ka\rbrace)}{n}-\int_{[0,1]}f$, then we have
$d_n=\displaystyle\sum_{k=1}^{2^q} d_{n,j}$ where
$$
d_{n,j}=\frac{\displaystyle\sum_{k\in X_{n,j}}f(\lbrace ka \rbrace)}{n}-
\int_{I_j} f \tag{4}
$$
On each interval $I_j$, $f$ is continuous and so attains a minimum
value $m_j$. We then have
$$
\begin{array}{lclc}
|d_{n,j}| & \leq &
\Bigg|
{\displaystyle\sum_{k\in X_{n,j}}\frac{f(\lbrace ka \rbrace)-m_j}{n}}
\Bigg|+
\Bigg|
{\displaystyle\sum_{k\in X_{n,j}}\frac{m_j}{n}-\int_{I_j}m_j}
\Bigg|+
\Bigg|
\int_{I_j}(f-m_j)
\Bigg|
\\
& = &
\Bigg|
{\displaystyle\sum_{k\in X_{n,j}}\frac{f(\lbrace ka \rbrace)-m_j}{n}}
\Bigg|+
|m_j|\Bigg|f_n(I_j)-\frac{1}{2^q}\Bigg|+
\Bigg|
\int_{I_j}(f-m_j)
\Bigg|
\\
& \leq &
{\displaystyle\sum_{k\in X_{n,j}}\frac{|f(\lbrace ka \rbrace)-m_j|}{n}}
+M\eta+
\int_{I_j}|f-m_j|
\\
& \leq &
{\displaystyle\sum_{k\in X_{n,j}}\frac{\eta}{n}}
+M\eta+
\int_{I_j}\eta
\\
& \leq &
\eta
+M\eta+
\frac{\eta}{2^q}=\eta(M+1+\frac{1}{2^q})
\\
\end{array}
$$
Summing on $j$, we deduce
$$
|d_n| \leq \eta (2^q(M+1)+1)
$$
Taking $\eta=\frac{\varepsilon}{2^q(M+1)+1}$, we have $|d_n| \leq \varepsilon$,
so we have shown that $(d_n)$ tends to zero as wished.
