Here is an alternative solution to the OP based in continued fractions. These types of fractions yield the best rational approximations to numbers. First, a few general facts about continued fractions:
For simplicity, I only focus on numbers in the interval $[0,1]$. Define $$T(x)=\left\{\begin{array}{lcr}
\frac{1}{x}-\lfloor \frac1x\rfloor &\text{if} & x\neq0\\
0 &\text{if} & x=0
\end{array}\right.$$
where $\lfloor \cdot\rfloor$ is the floor function (a.k.a. maximum integer function).
Then, if $0<x<1$
$$x=\frac{1}{1/x}=\frac{1}{\lfloor \frac1x\rfloor + Tx}=\frac{1}{a_1 +Tx}$$
where $a_1(x)=\lfloor \frac1x\rfloor$. Continuing this way, and as long as $T^{n-1}x=T(T^{n-2}(x))\neq0$, let $a_n(x)=\lfloor \tfrac{1}{T^{n-1}x}\rfloor$. We obtain the sequences of rational numbers
$$ x_n:=\frac{1}{a_1+\tfrac{1}{a_2 +\ddots\tfrac{1}{a_{n-1}+\tfrac{1}{a_n}}}}$$
It turns out that the rational approximations $x_n$ in some sense that can be made very precise:
- If $x$ is rational, the sequence $x_n$ is final after a finite number of steps.
For irrational $x$, the sequence $x_n$ is infinite and
$x_n=\frac{P_n}{Q_n}$ where
\begin{align}
P_n&=a_nP_{n-1}+P_{n-2}\tag{1}\label{one}\\
Q_n&=a_nQ_{n-1}+Q_{n-2}\tag{2}\label{two}
\end{align}
with $P_0=0$, $P_1=1$, $Q_0=1$ and $Q_1=a_1$.
It is relatively easy to check that $g.c.d(P_n,Q_n)=1$ for all $n\geq0$, i.e. all fractions $P_n/Q_n$ are reduced.
With a little more effort we have the bounded
\begin{align}
\frac{1}{Q_nQ_{n+2}}<\Big|x-\frac{P_n}{Q_n}\Big|<\frac{1}{Q_nQ_{n+1}}\tag{3}\label{three}
\end{align}
Solution to the OP:
That the rational $x_n$'s are the best rational approximations to $x$ is expressed in the following result
Theorem A: If there is a rational number $a/b$ with $b>0$ such that
\begin{align}\big|x-\frac{a}{b}\big|<\big|x-\frac{P_n}{Q_n}\big|\tag{4}\label{four}\end{align}
for some $n>0$, then $b>Q_n$.
If $x$ is irrational, \eqref{one} and \eqref{two} imply that both $P_n(x), Q_n(x)$ are strictly monotone and thus, $P_n,Q_n\xrightarrow{n\rightarrow\infty}\infty$. By \eqref{three} $\frac{P_n}{Q_n}\xrightarrow{n\rightarrow\infty}x$. The conclusion of the OP follows from \eqref{four}.
Comment: It is customary to use the notation $[a_1a_1\ldots]$ for the number $x$; other notations are $\frac{1}{a_1+}\frac{1}{a_2+}\cdots$.