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Let

• 
  $(Omega,mathcal A,operatorname P)$ be a probability space


• 
  $(mathcal F_t)_{tge0}$ be a filtration on $(Omega,mathcal A)$
  


• 
  $E$ be a metric space


• 
  $(X_t)_{tge0}$ be an $E$-valued right-continuous time-homogeneous Markov process on $(Omega,mathcal A,operatorname P)$
  


• 
  $kappa_t$ be a regular version of the conditional probability of $X_t$ given $X_0$, i.e. $kappa_t$ is a Markov kernel on $(E,mathcal B(E))$ with $$operatorname Pleft[X_{s+t}in Bmidmathcal F_sright]=operatorname Pleft[X_tin Bmid X_0right]=kappa_t(X_0,B);;;text{almost surely for all }Binmathcal B(E)tag1$$ for all $s,tge0$
  


• 
  $f:Etomathbb R$ be bounded and continuous

Are we able to show that $$[0,infty)ni tmapsto(kappa_t f)(x):=intkappa_t(x,{rm d}y)f(y)tag1$$ is continuous for all $xinmathbb R$?

Let's take a look: Let $(t_n)_{ninmathbb N}subseteq[0,infty)$ and $tge0$ with $t_nxrightarrow{ntoinfty}t$. For simplicity, assume $t=0$. By the dominated convergence theorem, $$operatorname Eleft[fleft(X_{t_n}right)mid X_0right]xrightarrow{ntoinfty}operatorname Eleft[fleft(X_0right)mid X_0right]=f(X_0);;;text{almost surely}tag2$$ and hence $$operatorname Eleft[fleft(X_{t_n}right)mid X_0=;cdot;right]xrightarrow{ntoinfty}operatorname Eleft[f(X_0)mid X_0=;cdot;right]=f;;;operatorname Pcirc:X_0^{-1}text{-almost surely}.tag3$$ By $(1)$, $$operatorname Eleft[fleft(X_{t_n}right)mid X_0=;cdot;right]=kappa_{t_n}f;;;text{for all }ninmathbb N;operatorname Pcirc:X_0^{-1}text{-almost surely}.tag4$$ Thus, $$kappa_{t_n}fxrightarrow{ntoinfty}f;;;operatorname Pcirc:X_0^{-1}text{-almost surely}.tag4$$

The problem with $(4)$ is the dependence of the null set on $(t_n)_{ninmathbb N}$.

However, $left|kappa_{t_n}right|leleft|fright|_infty$ for all $ninmathbb N$ and hence another application of the dominated convergence theorem yields $$operatorname Eleft[left(kappa_{t_n}fright)(X_0)right]xrightarrow{ntoinfty}operatorname Eleft[f(X_0)right]tag5.$$

If we fix an $xin E$ and assume $operatorname Pcirc:X_0^{-1}=delta_x$, the continuity of $(1)$ is obvious from $(5)$.

Now, Mars Plastic wrote in his answer that the assumption $operatorname Pcirc:X_0^{-1}=delta_x$ is no restriction.

Why?

I have some crazy idea: Assume $E$ is separable. Let $D([0,infty),E)$ denote the space of càdlàg functions $[0,infty)to E$ equipped with the Skorohod topology and $$pi_t:D([0,infty),E)to E;,;;;xmapsto x(t)$$ for $tge0$. By separability, $$mathcal Bleft(D([0,infty),E)right)=sigma(pi_t:tge0)tag6.$$ Now, assume that there is a Markov kernel $kappa$ with source $(E,mathcal E)$ and target $$(tilde Omega,tilde{mathcal A}):=left(D([0,infty),E),mathcal Bleft(D([0,infty),E)right)right)$$ with $$kappa(x,;cdot;)circleft(pi_{t_0},ldots,pi_{t_n}right)^{-1}=delta_xotimesbigotimes_{i=1}^nkappa_{t_i-t_{i-1}};;;text{for all }ninmathbb Ntext{ and }0=t_0<cdots<t_ntag7$$ for all $xin E$. Then, if $mu$ is any probability measureon $(E,mathcal E)$ and $$tilde{operatorname P}[tilde A]:=(mukappa)(tilde A)=intmu({rm d}x)kappa(x,tilde A);;;text{for }tilde Aintilde{mathcal A},$$ then it's easy to see that $(pi_t)_{tge0}$ is a càdlàg Markov process on $(tildeOmega,tilde{mathcal A},tilde{operatorname P})$ with transition semigroup $(kappa_t)_{tge0}$ and initial distribution $operatorname Pcirc:pi_0^{-1}=mu$.

Clearly, we could choose $mu=delta_x$ for some fixed $xin E$ and immediately obtain the continuity of $(1)$ by the same argumentation as before.

First of all, I have no idea if we can prove the existence of $kappa$. As you may guess, my idea was inspired by the usual existence proof of a Markov process with a given transition semigroup and initial distribution. Therein, $E$ is assumed to be a Polish space, $pi_t$ is replaced by $E^{[0,:infty)}ni xmapsto x(t)$ and $(tildeOmega,tilde{mathcal A})$ is replaced by $left(E^{[0,:infty)},mathcal E^{otimes[0,:infty)}right)$. The definition of $tilde{operatorname P}$ is the same.

So, if this is the correct approach, the concrete Markov process $(X_t)_{tge0}$ in the question is of no use. All we need is the transition semigroup $(kappa_t)_{tge0}$ satisfying the usual consistency conditions (Chapman-Kolmogorov equations) and then hope that $E$ is "nice enough" to admit the existence of $kappa$.

So, the question is: Is this the correct approach? And if so: When is $E$ is "nice enough"?

Your presentation is slightly bulky, but the general argument is correct. Note that you can fix a point $xin mathbb R$ and that it suffices to consider $Pcirc X_0^{-1}=delta_{x}$. Then your reasoning yields that for any $(t_n)subsetmathbb [0,infty)$ converging to $tin[0,infty)$ we have
begin{equation}
(kappa_{t_n}f)(x)=E[f(X_{t_n})|X_0=x] to E[f(X_t)|X_0=x]=(kappa_tf)(x), quad ntoinfty,
end{equation}
and that's all you need. There are no null sets which depend on anything that bothers us.

PS: Since you are in a time-homogeneous setting, it is indeed irrelevant if you consider $t=0$ or any other $tin [0,infty)$.