Show that $∀ x_1, x_2∈ (0,1)$, there exists integers $1 \leq q \leq 100$ and $0\leq p_1, p_2 \leq q$ such that $|x_1-\frac{p_1}{q}| \leq \frac{1}{10q}$ and $|x_2-\frac{p_2}{q}| \leq \frac{1}{10q}$
I've tried applying Dirichlet's approximation theorem separately to $x_1$ and $x_2$ to see if I can manipulate the inequalities to find a common $q$, but that approach doesn't generate a small enough bound on the approximation. I'm thinking that the proof has a similar construction to the proof using the pigeonhole principle for the Dirichlet's approximation theorem, but I am not entirely sure how to proceed.
Cut the unit square into $100$ little squares, each of side $1/10$, in the obvious way. Write $\{x\}$ for the fractional part of $x$. Consider the $101$ points $(\{jx_1\},\{jx_2\})$, $0\le j\le100$. Two of them, say $(\{mx_1\},\{mx_2\})$ and $(\{nx_1\},\{nx_2\})$, $0\le m<n\le100$, must be in the same little square. So $\{qx_1\}\le1/10$ and $\{qx_2\}\le1/10$, where $q=n-m$; note $1\le q\le100$. So there are integers $p_1$ and $p_2$ such that $|qx_1-p_1|\le1/10$ and $|qx_2-p_2|\le1/10$. Now divide through by $q$.