In my reference the theorem is stated as, convergents to an (irrational) number give a sequence of best approximations.
It is also given a proof by contradiction :
Assuming $|q\alpha-p|<|q_n\alpha-p_n|$ and $q\le q_n$
Consider the set of equations $$ xp_n+yp_{n+1}=p\\ xq_n+yq_{n+1}=q\\ $$ which can be written as $$ \begin{bmatrix}p_n&p_{n+1}\\q_n&q_{n+1}\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}p\\q\end{bmatrix} $$ where $\begin{vmatrix}p_n&p_{n+1}\\q_n&q_{n+1}\end{vmatrix}=p_nq_{n+1}-p_{n+1}q_n=(-1)^n\neq 0$ $$ \implies \begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}p_n&p_{n+1}\\q_n&q_{n+1}\end{bmatrix}^{-1}\begin{bmatrix}p\\q\end{bmatrix}=\frac{1}{(-1)^n}\begin{bmatrix}q_{n+1}&-p_{n+1}\\-q_n&p_n\end{bmatrix}\begin{bmatrix}p\\q\end{bmatrix}=\begin{bmatrix}pq_{n+1}-qp_{n+1}\\-pq_n+qp_n\end{bmatrix} $$ Therefore, the pair of equations has integer solutions $x,y$.
Case 1 : $y=0\implies p=xp_n$ and $q=xq_n$, $x\neq0$
$q\alpha-p=|x||q_n\alpha-p_n|\geq|q_n\alpha-p_n|$
Case 2 : $x=0$ and $y\neq 0$ then $q=yq_{n+1}\implies q>q_{n+1}>q_n$ since $q,q_n>0$, which is a contradiction to $q\leq q_n$.
Case 3 : $x$ and $y$ cant both be less than $0$ as $q>0$
Case 4 : If both $x$ and $y$ are positive then $q>q_{n+1}>q_n$ which again is a contradiction to $q\leq q_n$.
This concludes that $x$ and $y$ are of opposite sign.
Since $\alpha$ lies between two alternative convergents, $q_n\alpha-p_n$ and $q_{n+1}\alpha-p_{n+1}$ have opposite signs.
$$ q\alpha-p=x(q_n\alpha-p_n)+y(q_{n+1}\alpha-p_{n+1})\\ |q\alpha-p=|x(q_n\alpha-p_n)+y(q_{n+1}\alpha-p_{n+1})|=\\ |x(q_n\alpha-p_n)|+|y(q_{n+1}\alpha-p_{n+1})|>|x(q_n\alpha-p_n)|\geq |(q_n\alpha-p_n)| $$ which contradicts the assumption and proves the theorem.
The statement of the theorem says $\alpha$ is an irrational number, but I do not see anywhere in the proof we take $\alpha$ as an irrational number, so do the theorem and the above proof applies to all real numbers ? Or am I missing something here ?