= Solution
Write the interval as $I=(c,d]$, where $0<c<d<1$, to avoid confusing its left endpoint with the rotation parameter $\alpha$. For every sufficiently large integer $k$, define an inner and outer interval on the circle by
$$
I_k^-=(c+k^{-1},d-k^{-1}],
\qquad
I_k^+=(c-k^{-1},d+k^{-1}].
$$
Apply part (b) simultaneously to this countable collection of intervals. The intersection $G$ of the corresponding full-measure sets still has full <Lebesgue measure>, and is therefore <dense subset>[dense] in the circle.
Fix an arbitrary $x$. For each sufficiently large $k$, choose $y_k\in G$ with circle distance less than $1/k$ from $x$. An <irrational rotation of the circle>[irrational circle rotation] preserves this distance, so for every $j\geq0$,
$$
\mathbf1_{I_k^-}(T^jy_k)
\leq\mathbf1_I(T^jx)
\leq\mathbf1_{I_k^+}(T^jy_k).
$$
Averaging and using part (b) for $y_k$ gives
$$
d-c-\frac2k
\leq
\liminf_{n\to\infty}\frac{S_n(\mathbf1_I)(x)}n
\leq
\limsup_{n\to\infty}\frac{S_n(\mathbf1_I)(x)}n
\leq
d-c+\frac2k.
$$
Letting $k\to\infty$ proves the <everywhere interval frequency under an irrational rotation>:
$$
\boxed{
\frac{S_n(\mathbf1_{(c,d]})(x)}n\longrightarrow d-c
\quad\text{for every }x\in(0,1].}
$$
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