We have the standard Dirichlet approximation theorem that states that, for $\alpha \in (0,1)$,
$$ \min_{1 \leqslant q \leqslant n} q \alpha \,(\text{mod } 1)\, \xrightarrow{unif.} 0 $$
uniformly in $\alpha$.
I want to show that this is also true when $q$ is replaced with $q^2$ - i.e.
$$ \min_{1 \leqslant q \leqslant n} q^2 \alpha \,(\text{mod } 1) \, \longrightarrow 0 $$
Is this convergence also uniform in $\alpha$?
Unfortunately, this no longer seems to be possible to prove with a simple application of the pigeonhole principle.
However, I have been given a hint to use Roth’s theorem on arithmetic progressions (that any subset of $\{1,...,n\}$ with density at least $\frac{Cn}{\log\log n}$ has a three term arithmetic progression).
First of all, the usual phrasing of the Dirichlet approximation theorem replaces "$(\text{mod } 1)$" with $\| \cdot \|_\mathbb{T}$, which gives the distance to the nearest integer (i.e. the distance to $0$ in $\mathbb{T} = \mathbb{R}/\mathbb{Z}$), so $\| 1\frac34 \|_\mathbb{T} = \frac14$, whereas $1\frac34 \pmod{1}$ could mean $\frac34$. This is what is proved using the pigeonhole principle in the Wikipedia article you reference.
We can actually use a similar strategy to the proof of the Dirichlet approximation theorem.
Let $k \in \mathbb{N}$, and consider the sets
$$S_i := \left\{ q \in \{1, ..., n\} \, : \, \left \| \frac{q^2 \alpha}{2} \, \right\|_\mathbb{T} \in \left[\frac{i}{k}, \frac{i+1}{k} \right) \right\}$$
for $i = 0, ..., k-1$.
The sets, $S_i$, partition $\{1, ..., n\}$, so by the pigeonhole principle, there is a set, $S = S_j$, containing at least $\frac{n}{k}$ elements.*
If we take $n$ large enough (i.e. so that $\frac {C}{\log \log n} \leqslant \frac1k$), then Roth’s theorem gives us an arithmetic progression $\{x-d, x, x+d\} \subseteq S$.
Then
\begin{align} \|d^2 \alpha\|_\mathbb{T} &= \left \|\frac{(x-d)^2 \alpha}{2} + \frac{(x+d)^2 \alpha}{2} - x^2 \alpha \right\|_\mathbb{T} \\ &\leqslant \left \| \frac{(x-d)^2 \alpha}{2} - \frac{x^2 \alpha}{2} \right \|_\mathbb{T} + \left \| \frac{(x+d)^2 \alpha}{2} - \frac{x^2 \alpha}{2} \right \|_\mathbb{T} \\ &\leqslant \frac1k + \frac1k = \frac2k \end{align}
and so
$$\min_{1 \leqslant q \leqslant n} \| q^2 \alpha \|_\mathbb{T} \leqslant \frac2k$$
Since $k$ was arbitrary, we are done.
Note that $n$ only needed to be large enough in terms of $k$ - it was chosen independently of $\alpha$, so convergence is indeed uniform in $\alpha$.
*In the proof of the Dirichlet approximation theorem, at this point, as long as $n > k$, we have two distinct elements $q_1\alpha$, $q_2\alpha$ that are at most $2/k$ apart, modulo 1. Taking their difference, $Q = q_2 - q_1$, then gives $Q \alpha\pmod{1} \leqslant 2/k$ and $1 \leqslant Q \leqslant k+1$.
However, we can't quite do the same think in this case. If we set $Q = q_1^2 - q_2^2$, then whilst $Q \alpha \leqslant 2/k$, $Q$ need not be a square, so is not in the required form. This is why we need to use Roth's theorem - if we have three squares in arithmetic progression which are close together modulo 1, we can form a new square which is close to zero, modulo 1.