Limit associated with a recursion, connection to normality of quadratic irrationals

The proof follows rather easily from all the data you gathered. As often with recurrences, the core idea is to realize that if the conjecture holds for $y_0,z_0$ it holds for $y_n,z_n$ as well, and deduce new, non-trivial consequences from that.

Generalizing your formula for $x$, let us put

$$ x_n =\frac{-(z_n-1)+\sqrt{(z_n-1)^2+8y_n}}{4} \tag{1} $$

As you already computed, $x_n$ is a root of $P_n=x^2 +(z_n-1)x -y_n$. If your conjecture is correct (and it is, as will be shown soon), $x_n$ should be in $[0,1]$. It turns out that this is true because of the form of $P_n$.

Lemma 1. For every $n$, $P_n$ is increasing on $[0,1]$, and satisfies $P_n(0) \lt 0 \lt P_n(1)$.

Proof of lemma 1 : Since $P'_n(0)=z_n-1$, $P_n(0)=1-y_n$ and $P_n(1)=z_n+1-y_n$, it suffices to show that $z_n\geq 1,1-y_n\lt 0 \lt z_n+1-y_n$ for all $n$. This is straightforward by induction on $n$.

Let $\delta_n$ be the second digit in the dyadic expansion of $x_n$ (it will soon turn out that $\delta_n$ is the same thing as your $d_{n+1}$). We wish to know if $\delta_n$ is zero or $1$, in other words whether $x_n$ is smaller or larger than $\frac{1}{2}$, or what is the sign of $P_n(\frac{1}{2})$.

But

$$ P_n(\frac{1}{2})=\frac{z_n-2y_n}{2} \tag{2} $$

Now you know where your comparison of $z_n$ to $2y_n$ comes from ! (2) also shows that $\delta_n=d_{n+1}$. Furthermore, a purely algebraic verification shows that the recursion on $y_n$ and $z_n$ is equivalent to

$$ P_{n+1}(x)=4P_{n}\bigg(\frac{\delta_n + x}{2}\bigg) \tag{3} $$

Finally, it follows from (3) that

$$ x_n=\frac{\delta_n+x_{n+1}}{2} \tag{4} $$

and hence

$$ x_n=\sum_{j=n}^{\infty} \frac{\delta_j}{2^j} \tag{5} $$

This finishes the proof.