[proofplan]
The length of the interval is the absolute value of the difference between two consecutive convergents. We compute that difference by proving the determinant identity $p_{n+1}q_n - p_n q_{n+1} = (-1)^n$, which follows directly from the recurrence defining the convergents. Dividing this determinant identity by the positive product $q_n q_{n+1}$ gives the signed difference, and taking absolute values gives the stated length.
[/proofplan]
[step:Prove the determinant identity for consecutive convergents]
For each integer $n \geq 0$, define the integer
\begin{align*}
D_n := p_{n+1}q_n - p_n q_{n+1}.
\end{align*}
We prove that
\begin{align*}
D_n = (-1)^n
\end{align*}
for every $n \geq 0$.
For $n = 0$, the recurrence gives
\begin{align*}
p_0 = a_0, \qquad q_0 = 1.
\end{align*}
It also gives
\begin{align*}
p_1 = a_1p_0 + p_{-1} = a_1a_0 + 1, \qquad q_1 = a_1q_0 + q_{-1} = a_1.
\end{align*}
Therefore
\begin{align*}
D_0 = p_1q_0 - p_0q_1 = (a_1a_0 + 1)\cdot 1 - a_0a_1 = 1 = (-1)^0.
\end{align*}
Now let $n \geq 1$ and assume the identity has been established for $D_{n-1}$. Using the recurrence for $p_{n+1}$ and $q_{n+1}$, we compute
\begin{align*}
D_n = p_{n+1}q_n - p_nq_{n+1}.
\end{align*}
Substituting the recurrence formulas for $p_{n+1}$ and $q_{n+1}$ gives
\begin{align*}
D_n = (a_{n+1}p_n + p_{n-1})q_n - p_n(a_{n+1}q_n + q_{n-1}).
\end{align*}
Expanding the products and cancelling the two identical terms $a_{n+1}p_nq_n$ gives
\begin{align*}
D_n = p_{n-1}q_n - p_nq_{n-1}.
\end{align*}
By the definition of $D_{n-1}$, this is
\begin{align*}
D_n = -(p_nq_{n-1} - p_{n-1}q_n) = -D_{n-1}.
\end{align*}
By the induction hypothesis, $D_{n-1} = (-1)^{n-1}$, so
\begin{align*}
D_n = -(-1)^{n-1} = (-1)^n.
\end{align*}
Thus $p_{n+1}q_n - p_nq_{n+1} = (-1)^n$ for every $n \geq 0$.
[guided]
The quantity that controls the difference between consecutive fractions is the cross-determinant. For each $n \geq 0$, define
\begin{align*}
D_n := p_{n+1}q_n - p_nq_{n+1}.
\end{align*}
This is the numerator that appears when the two fractions $p_{n+1}/q_{n+1}$ and $p_n/q_n$ are placed over the common denominator $q_nq_{n+1}$. We will prove
\begin{align*}
D_n = (-1)^n
\end{align*}
for every $n \geq 0$.
First consider $n = 0$. From the initial conditions and the recurrence,
\begin{align*}
p_0 = a_0p_{-1} + p_{-2} = a_0, \qquad q_0 = a_0q_{-1} + q_{-2} = 1.
\end{align*}
The next convergent is determined by
\begin{align*}
p_1 = a_1p_0 + p_{-1} = a_1a_0 + 1, \qquad q_1 = a_1q_0 + q_{-1} = a_1.
\end{align*}
Substituting these four values into the definition of $D_0$ gives
\begin{align*}
D_0 = p_1q_0 - p_0q_1 = (a_1a_0 + 1)\cdot 1 - a_0a_1 = 1 = (-1)^0.
\end{align*}
So the identity holds at the initial index.
Now suppose $n \geq 1$. The reason to use the determinant $D_n$ is that the terms involving $a_{n+1}$ cancel exactly. Applying the recurrence formulas
\begin{align*}
p_{n+1} = a_{n+1}p_n + p_{n-1}, \qquad q_{n+1} = a_{n+1}q_n + q_{n-1},
\end{align*}
we obtain
\begin{align*}
D_n = p_{n+1}q_n - p_nq_{n+1}.
\end{align*}
After substitution,
\begin{align*}
D_n = (a_{n+1}p_n + p_{n-1})q_n - p_n(a_{n+1}q_n + q_{n-1}).
\end{align*}
Expanding the products gives
\begin{align*}
D_n = a_{n+1}p_nq_n + p_{n-1}q_n - a_{n+1}p_nq_n - p_nq_{n-1}.
\end{align*}
Cancelling the two opposite copies of $a_{n+1}p_nq_n$ leaves
\begin{align*}
D_n = p_{n-1}q_n - p_nq_{n-1}.
\end{align*}
The remaining expression is the negative of the previous determinant:
\begin{align*}
p_{n-1}q_n - p_nq_{n-1}
=
-(p_nq_{n-1} - p_{n-1}q_n)
=
-D_{n-1}.
\end{align*}
Hence
\begin{align*}
D_n = -D_{n-1}.
\end{align*}
Since $D_0 = 1$, repeated sign reversal gives
\begin{align*}
D_n = (-1)^n
\end{align*}
for every $n \geq 0$. Therefore
\begin{align*}
p_{n+1}q_n - p_nq_{n+1} = (-1)^n
\end{align*}
for every $n \geq 0$.
[/guided]
[/step]
[step:Convert the determinant identity into the length formula]
Fix $n \geq 0$. Since $q_n > 0$ and $q_{n+1} > 0$, the product $q_nq_{n+1}$ is positive. Using the determinant identity from the previous step, we compute
\begin{align*}
\frac{p_{n+1}}{q_{n+1}} - \frac{p_n}{q_n} = \frac{p_{n+1}q_n - p_nq_{n+1}}{q_nq_{n+1}} = \frac{(-1)^n}{q_nq_{n+1}}.
\end{align*}
Taking absolute values and using $q_nq_{n+1} > 0$, we get
\begin{align*}
\left|\frac{p_{n+1}}{q_{n+1}} - \frac{p_n}{q_n}\right| = \left|\frac{(-1)^n}{q_nq_{n+1}}\right| = \frac{1}{q_nq_{n+1}}.
\end{align*}
The length of the closed interval in $\mathbb{R}$ with endpoints $p_n/q_n$ and $p_{n+1}/q_{n+1}$ is exactly the absolute difference of its endpoints. Therefore the interval has length
\begin{align*}
\frac{1}{q_nq_{n+1}}.
\end{align*}
[/step]
