We take, without future mention,
- $K$ to be an arbitrary set (not necessarily in $\mathbb{R}^n$),
- $g$ to be a bounded and nonnegative real-valued map from $K$ (not necessarily continuous),
- $f$ to be a map from $K$ to itself (not necessarily continuous), and
- $(\beta_n)_n$ to be a real sequence converging monotonically from below to one.
In particular, no topological structure is imposed on $K$.
Remark.
With regards to your original post, I don't see how $T_{\beta}$ is necessarily a map from $C(K)$ to itself since you impose no continuity on $f$.
Therefore, in stating the main result below, I use the space of bounded real-valued maps $B(K)$.
Theorem. Suppose the following:
(H1) For each $x$, the supremum $\sup_{k\geq0}[g\circ f^{k}](x)$ is attained at a nonnegative integer $k$.
(H2) The operator $T_{\beta}$ defined by $T_{\beta}F\equiv\max_{0 \leq a \leq 1} ag+(1-a)\beta[F\circ f]$ has a unique fixed point in $B(K)$ for each $\beta$ between zero and one (inclusive).
Denote by $F_{\beta}$ the unique fixed point of $T_{\beta}$.
Then, $F_{\beta_{n}}\rightarrow F_{1}$ pointwise.
We prove the above in a series of lemmas.
Conventions.
We always use $\beta$ to mean a number between zero and one (inclusive).
We take $\inf \emptyset = \infty$.
$\Vert \cdot \Vert$ denotes the sup-norm.
$f^k$ denotes the composition of $f$ with itself $k$ times.
First, since the maximum of a linear function on an interval is obtained at one of the boundaries,
$$
T_{\beta}F=\max_{0\leq a\leq1}ag+\left(1-a\right)\beta[F\circ f]=\max\left\{ g,\beta[F\circ f]\right\} .
$$
This suggests that the Bellman equation is the dynamic program associated to a deterministic optimal stopping problem in which the state at time $k$ is $X_{k}\equiv f^{k}(x)$.
The controller has to choose a time $k$ to stop at to receive the reward $\beta^{k}g(X_{k})=\beta^{k}[g\circ f^{k}](x)$.
You can think of $(X_k)_k$ as a Markov chain with deterministic transitions.
Unless there is no discount ($\beta=1$), the controller prefers to stop earlier in order to minimize adverse discounting effects.
This is made rigorous below.
Lemma 1 (Characterizing a fixed point). $V_{\beta}\equiv\sup_{k\geq0}\beta^{k}[g\circ f^{k}]$
is a fixed point of $T_{\beta}$.
Proof. Substituting $V_{\beta}$ into $T_{\beta}$,
$$
T_{\beta}V_{\beta}=\max\left\{ g,\sup_{k\geq0}\beta^{k+1}[g\circ f^{k+1}]\right\} =\sup_{k\geq0}\beta^{k}[g\circ f^{k}]
=V_{\beta}.
$$
Lemma 2 (Properties of $V_{\beta}$). $V_{\alpha}\leq V_{\beta}$ whenever $0\leq\alpha\leq\beta$ and $\sup_{\beta}\Vert V_{\beta}\Vert<\infty$.
Proof. The first inequality follows immediately from the definition of $V_{\beta}$ in Lemma 1.
As for the second inequality, note that $\beta^{k}\leq1$ and hence $\Vert V_{\beta}\Vert\leq\Vert g\Vert < \infty$.
Corollary 3 (Convergence). The (pointwise) limit $V\equiv\lim_n V_{\beta_{n}}$ exists.
Proof. Apply the monotone convergence theorem pointwise to $V_{\beta_{n}}(x)$.
Next, we define the optimal stopping time
$$
k_{\beta}(x)\equiv\inf\left\{ k\geq0\colon V_{\beta}(x)=\beta^{k}[g\circ f^{k}](x)\right\} .
$$
Since $g$ is nonnegative and $\beta<1$, it follows that
$$
0\leq \beta^{k}[g\circ f^{k}](x)\leq\beta^{k}\left\Vert g\right\Vert \rightarrow0\text{ as }k\rightarrow\infty
$$
and hence $k_{\beta}<\infty$.
However, establishing $k_1 < \infty$ requires assuming (H1).
Lemma 4. $V\leq V_{1}$.
Proof. Let $x$ be arbitrary.
Letting $k_{n}\equiv k_{\beta_{n}}$ for brevity,
\begin{multline*}
V(x)
=\lim_{n}V_{\beta_{n}}(x)
=\lim_{n}\left(\beta_{n}\right)^{k_{n}(x)}[g\circ f^{k_{n}(x)}](x)\\
\leq\lim_{n}[g\circ f^{k_{n}(x)}](x)
\leq\sup_{k\geq0}[g\circ f^{k}](x)
=V_{1}(x).
\end{multline*}
Lemma 5. Suppose (H1). Then, $V\geq V_{1}$.
Proof. Let $x$ be arbitrary. Then,
\begin{multline*}
V(x)
=\lim_{n}V_{\beta_{n}}(x)
=\lim_{n}\sup_{k\geq0}\left(\beta_{n}\right)^{k}[g\circ f^{k}](x)\\
\geq\lim_{n}\left(\beta_{n}\right)^{k_{1}(x)}[g\circ f^{k_{1}(x)}](x)
=[g\circ f^{k_{1}(x)}](x)
=V_{1}(x).
\end{multline*}
We are now ready to prove the theorem.
Proof of the Theorem. Since fixed points are unique by assumption (H2), $V_{\beta}=F_{\beta}$, and the result follows from combining Lemmas 4 and 5.
It is worthwhile, as a closing note, to discuss assumption (H1).
Intuitively, (H1) tells us that the controller can always choose a finite optimal stopping time (this does not preclude the existence of infinite optimal stopping times).
There are some interesting cases that trivially satisfy (H1), such as (i) $K$ being a finite set or (ii) the image of $K$ under $g$ being a finite set.
Best Answer
As I was not able to show with Blackwell's sufficient conditions or directly that the operator T constitutes an contraction mapping, I rewrote the problem, to attack it with different methods. Define the function $g= h + f$ and add $h$ to each side of the equation above. Then, define the operator $\tilde{T}$ as:
$\tilde{T}g(x) = h(x) + \beta \left\lbrace \sum_{\theta} \mu_{\theta} \left[ \int g(x') Q_\theta (x, dx') \right]^{\alpha} \right\rbrace^{1/\alpha}$
Now, Marinacci and Montrucchio in their paper titled ''Unique solutions for stochastic recursive utilities'' operators of the folllwing form: $\hat{T} g = W \left( h,\mathcal{M}(g) \right) $ with with $W: \mathbf{R}^{2}_{+} \rightarrow \mathbf{R}_{+}$ and $\mathcal{M}: C(X) \rightarrow C(X) $. In particular, they lay out sufficient conditions for the operator $\hat{T}$ to be a contraction mapping. I then showed that my problem satifies these conditions and that I can formulate my problem through the function $W$ and operator $\mathcal{M}$. Having found a solution $g^{\star}$, I can retrieve the function $f^{\star}$.