= Solution
There is a minor range issue in the printed bound: \b[for $\varepsilon>1$ and positive $C$, the upper bound is below one.] We prove the intended result for $0<\varepsilon\leq1$; a valid formulation for every $\varepsilon>0$ replaces the upper bound by $\max\{1,\varepsilon^{-C}\}$. In fact the argument below gives $C=32$. All auxiliary estimates are proved here.
For $0<\varepsilon\leq1/2$, let $m=\lceil\varepsilon^{-2}\rceil$, and use the <Fejér kernel>
$$
F_m(t)=\frac1m\left|\sum_{j=0}^{m-1}e(jt)\right|^2
=\sum_{|k|<m}\left(1-\frac{|k|}{m}\right)e(kt).
$$
Expanding the square proves the identity and nonnegativity. The <finite geometric series> formula and $|\sin\pi t|\geq2\|t\|$ give $F_m(t)\leq(4m\|t\|^2)^{-1}$ off the <integers>. In particular $F_m(t)\leq1/4$ whenever $\|t\|\geq\varepsilon$.
Suppose, for a contradiction, that no $1\leq n\leq N$ has $\|\alpha n^2\|<\varepsilon$. Summing the <Fejér kernel> along the quadratic sequence and separating its constant term gives
$$
\frac34N\leq\sum_{0<|k|<m}\left(1-\frac{|k|}{m}\right)|S_k|,
\qquad S_k=\sum_{n=1}^N e(k\alpha n^2).
$$
The coefficients on the right sum to $m-1$. Also $|S_{-k}|=|S_k|$. Hence some $1\leq k<m$ has $|S_k|\geq\eta N$, where $\eta=1/(2m)$.
We need only an elementary <Van der Corput inequality for finite scalar sequences>. For $|z_n|\leq1$, extended by zero outside $[N]$, each term of $S=\sum_nz_n$ occurs in exactly $H$ windows of length $H$. Applying the <Cauchy-Schwarz inequality> to the window sums and expanding their squares gives, for $H\leq N$,
$$
H^2|S|^2\leq(N+H)\left(HN+2\sum_{h=1}^{H-1}(H-h)|C_h|\right),
\qquad C_h=\sum_{n=1}^{N-h}z_{n+h}\overline{z_n}.
$$
Consequently $|S|^2\leq2N^2/H+(4N/H)\sum_{h=1}^{H-1}|C_h|$. Take $H=\lceil8\eta^{-2}\rceil$. If $|S|\geq\eta N$, some $1\leq h<H$ must have $|C_h|\geq\eta^2N/8$; otherwise the displayed upper bound is less than $3\eta^2N^2/4$.
For the large <quadratic exponential sum> just found, put $z_n=e(k\alpha n^2)$. Its <quadratic exponential sum> has multiplicative derivative
$$
z_{n+h}\overline{z_n}=e(k\alpha h^2)e(2kh\alpha n).
$$
Thus $C_h$ is a <finite geometric series>. Its absolute value is at most $1/(2\|2kh\alpha\|)$, unless that distance is zero, in which case the desired estimate is automatic. It follows that
$$
\|2kh\alpha\|\leq\frac4{\eta^2N}.
$$
Set $n_*=2kh$. The <distance to the nearest integer> satisfies $\|\ell x\|\leq\ell\|x\|$ for a positive integer $\ell$, by multiplying a nearest integer to $x$. Therefore
$$
1\leq n_*\leq2mH,
\qquad
\|\alpha n_*^2\|\leq\frac{8mH}{\eta^2N}.
$$
This proves the needed <quantitative quadratic recurrence> once $N$ is chosen polynomially in $\varepsilon^{-1}$.
For explicit bookkeeping, $m\leq2\varepsilon^{-2}$ and $H\leq33m^2$. Choose $N=\lfloor\varepsilon^{-32}\rfloor$. For $0<\varepsilon\leq1/2$ we have $N>33792\varepsilon^{-11}\geq8mH/(\eta^2\varepsilon)$ and $N\geq H$, while $2mH\leq528\varepsilon^{-6}\leq N$. Hence $n_*\leq N$ and $\|\alpha n_*^2\|<\varepsilon$, contradicting our supposition. The estimates have substantial slack even at $\varepsilon=1/2$.
For $1/2<\varepsilon\leq1$, simply take $n=1$, since $\|\alpha\|\leq1/2<\varepsilon$ and $1\leq\varepsilon^{-32}$. \b[For the intended range, the conclusion is]
$$
\boxed{\exists\,1\leq n\leq\varepsilon^{-32}\text{ with }\|\alpha n^2\|<\varepsilon.}
$$
For $\varepsilon>1$, the same choice $n=1$ proves the corrected all-range version.
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