In the limit $d\to\infty$ the difference equation can be transformed into a ordinary differential equation, which can be solved: We define $\tau=t/d$ and expand 
$\beta(\tau) = \beta_{\tau d}$ around $d=\infty$,
\begin{align}\tag{1}\label{eq:1}
\beta(\tau) = \frac{4^{-\tau}(1-\tau \ln2)}{d} + \mathcal{O}(d^{-2}).
\end{align}
Now we rewrite the $x_t$-recursion also in terms of $\tau$ and get (with $\delta\tau=1/d$)
\begin{align}\tag{2a}\label{eq:2a}
x(\tau+\delta\tau)
&=\frac{x(\tau)+\sqrt{x(\tau)^2-4\beta(\tau)}}{2}\\
\tag{2b}\label{eq:2b}
&=\frac{x(\tau)+\sqrt{x(\tau)^2-4^{1-\tau}(1-\tau \ln2)\,\delta\tau+\mathcal{O}(\delta\tau^2)}}{2},
\end{align}
such that
\begin{align}
\tag{3a}\label{eq:3a}
x'(\tau)&=\lim_{\delta\tau\to 0}\frac{x(\tau+\delta\tau)-x(\tau)}{\delta\tau}\\
\tag{3b}\label{eq:3b}
&=-\frac{4^{-\tau}(1-\tau \ln2)}{x(\tau)}.
\end{align}
The solution of this ODE with initial condition $x(0)=1$ reads
\begin{align}
\tag{4}\label{eq:4}
x(\tau)=\sqrt{1-\frac{1}{\ln4}+4^{-\tau}\left(\frac{1}{\ln4}-\tau\right)}.
\end{align}
The convexity follows.