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Let $x=x(a,t)$ be a solution to the diff. eq.
\begin{equation*}
\dot{x}(a,t) = a - b\sin x(a,t). \tag{0}
\end{equation*}
The variables $t$ and $x$ are separable. Indeed, this diff. eq. can be rewritten as
\begin{equation*}
\int\frac{dx}{a - b\sin x}=\int dt,
\end{equation*}
whence
\begin{equation*}
\frac2\om\, \tan ^{-1}\frac{b-a \tan(\frac{x}{2})}{\om}=t+c
\end{equation*}
and
\begin{equation*}
x(a,t)=f(a,t+c), \tag{1}
\end{equation*}
for some real constant $c$, where
\begin{equation*}
f(a,t):=2 \tan ^{-1}\frac{b+\om \tan (\frac{1}{2}\om t)}{a}
\end{equation*}
and
\begin{equation*}
\om:=\om(a):=\sqrt{a^2-b^2}.
\end{equation*}
Letting
\begin{equation*}
T:=2\pi/\om, \tag{2}
\end{equation*}
note that $f(a,t)$ is smooth in $t\in(-T/2,T/2)$, with the partial derivative
\begin{equation*}
f'_t(a,t)=\frac{a\om^2}{a^2-b(b \cos \om t+\om \sin \om t)}\in\Big(\frac{\om^2}{a+b},\frac{\om^2}{a-b}\Big) \tag{3}
\end{equation*}
in $t$.
Also, $f(a,t)\to\pm\pi$ as $\pm t\uparrow T/2$.
Thus (and this is the crucial point), the function $(-T/2,T/2)\ni t\mapsto f(a,t)$ can be extended to a smooth function $\R\ni t\mapsto f(a,t)$ such that
\begin{equation*}
\psi(a,t):=f(a,t)-\om(a) t
\end{equation*}
is periodic in $t$ with period $T$ as in (2), and then (1) will define the general solution $x$ to the diff. eq. (0).
The graph of $f$ for $a=3,b=2.5$ is shown here:
Fix any real $x_0$. In view of (3), there is a unique real $c(a)$ such that
\begin{equation*}
f(a,c(a))=x_0, \tag{4}
\end{equation*}
and the the formula
\begin{equation*}
x(a,t)=f(a,t+c(a))=\om(a)t+\psi(a,t+c(a))
\end{equation*}
will define the unique solution to (0) satisfying the initial condition $x(a,0)=x_0$.
For each real $t$,
\begin{equation*}
x'_a(a,t)=\frac d{da}\,f(a,t+c(a))=\om'(a)t+\psi'_a(a,t+c(a))+\psi'_t(a,t+c(a))c'(a).
\end{equation*}
By (4), the implicit function theorem, and (3),
\begin{equation*}
|c'(a)|=|f'_a(a,c(a))/f'_t(a,c(a))|\ll|f'_a(a,c(a))|\ll\om'(a)|c(a)|+|\psi'_a(a,c(a))|;
\end{equation*}
the constants in $\ll$ may depend on $a$ and $b$, but not on $t$.
Since $\psi(a,t)$ is smooth in $(a,t)$ and periodic in $t$, it follows from the last two displays that
\begin{equation*}
|x'_a(a,t)|\ll t+1. \tag{5}
\end{equation*}
So, for $t>1$ we have $|x'_a(a,t)|\ll t$ and hence
\begin{equation*}
|x(a+\ep,t)-x(a,t)|\ll\ep t. \tag{6}
\end{equation*}
It remains to show that (6) holds for $t\in[0,1]$ as well. This is now easy: because of the smoothness of $f(a,t)$ in $(a,t)$ and the above bound on $|c'(a)|$, for $t\in[0,1]$
\begin{equation*}
|x(a+\ep,t)-x(a,t)|\le\int_0^t|f'_t(a+\ep,s+c(a+\ep))-f'_t(a,s+c(a))|\,ds\ll \int_0^t\ep\,ds=\ep t.
\end{equation*}
So, (6) holds for all $t\ge0$.
