Feller property for a collection of times Let $E$ be a locally compact Hausdorff space, $t\mapsto X_t$ a càdlàg Markov process on this space.
Suppose moreover that $X$ is Feller. What does Feller mean you ask? Well say that its semigroup $P_t$ maps $\mathcal C$ to $\mathcal C$, $\mathcal C$ being either the space of bounded continuous functions, or that of bounded continuous functions which vanish at infinity.1 I'm interested in both cases.
Fix some times $0<t_1\leq\cdots\leq t_n$ and a continuous function $F:E^n\to\mathbb R$ (maybe vanishing at infinity).

Is $x\mapsto\mathbb E_x[F(X_{t_1},\cdots,X_{t_n})]$ continuous?

In the case $n=1$, this is exactly the Feller property. It feels like one could prove this by induction once we know something like $x,\theta\mapsto\mathbb E_x[F_\theta(X_t)]$ is continuous, where $\theta,x\mapsto F_\theta(x)$ is continuous with $\theta$ an element of a suitable set $\Theta$ of parameters. This last property does not feel straightforward, though.
A reason for me to believe this might be true is the fact that it holds for diffusions with regular coefficients (on manifolds if you like). But I know no elementary proof of this fact; for what it's worth, it may be seen as an application of rough path theory.
1 A function $f:E\to\mathbb R$ vanishes at infinity if for any $\varepsilon>0$, there exists a compact $K\subset E$ over which $|f|<\varepsilon$.
 A: [All thanks and credit to saz in the comments.]
The answer is yes. Let's proceed step by step.
For $n\in\mathbb N^*$ and $t=(t_1,\cdots,t_n)$, define
$$P^n_t:F\mapsto\big(x\mapsto\mathbb E_x[F(X_{t_1},\cdots,X_{t_n})]\big)\text.$$

Suppose that $F:E^n\to\mathbb R$ is of the form $x_1,\cdots,x_n\mapsto g_1(x_1)\cdots g_n(x_n)$ with $g_i\in\mathcal C$.
Then $P^n_tF\in\mathcal C$.

The proof is by induction on $n$. The case $n=1$ is the Feller property. Now if the statement holds for $n-1$, then $\phi:x\mapsto\mathbb E_x[g_2(X_{t_2-t_1}),\cdots,g_n(X_{t_n-t_1})]$ must be in $\mathcal C$, and so would $x\mapsto g_1(x)\phi(x)$. Since $\mathbb E_x[F(X_{t_1},\cdots,X_{t_n})]=\mathbb E_x[g_1(X_{t_1})\phi(X_{t_1})]$ by the Markov property, we can apply the Feller property and conclude.

Suppose that $F:E^n\to\mathbb R$ is continuous and vanishes at infinity (with respect to the product topology).
Then $P^n_tF$ is continuous bounded.

Boundedness is clear. The continuity can be shown by an approximation argument. Let $A$ be the algebra of finite sums of functions of the form $x_1,\cdots,x_n\mapsto g_1(x_1)\cdots g_n(x_n)$ with $g_i$ continuous vanishing at infinity. According to the locally compact Stone-Weierstrass theorem, there exists a sequence $(G_k)_{k\geq0}$ of elements of $A$ such that $G_k\to F$ in the uniform sense. But this implies that $P^n_tG_k\to P^n_tF$ in the uniform sense, and since the former is continuous according to the previous claim, then the latter must be as well.

Suppose in addition that $E$ is second countable.
Then for any $F:E^n\to\mathbb R$ continuous bounded, $x\mapsto\mathbb E_x[F(X_{t_1},\cdots,X_{t_n})]$ is continuous bounded.

Again, boundedness is obvious. To show continuity, we fix $x\in E$ and show that $P^n_tF$ is continuous at $x$.
Fix $\varepsilon>0$. The first step is to find a suitable cutoff function to ignore what happens at infinity. Because $E$ is second countable locally compact, it is $\sigma$-compact. In particular, there exists a compact $K\subset E$ such that
$$\mathbb P_x(X_{t_i}\notin K\text{ for some }1\leq i\leq n)\leq\frac\varepsilon{3|F|_\infty}\text.$$
Because $E$ is locally compact Haussdorff, there exists a continuous function $\chi:E\to[0,1]$ with compact support such that $\chi\geq1_K$; it is a generalisation of Urysohn's lemma (for some reason, the generalisation appears on the French Wikipedia). By definition,
$$   \mathbb E_x[\chi(X_{t_1})\cdots\chi(X_{t_n})]
\geq \mathbb P_x[X_{t_1}\in K\text,\cdots\text,X_{t_n}\in K]
\geq 1-\frac\varepsilon{3|F|_\infty}\text.$$
But using the above result, we see that $y\mapsto \mathbb E_y[\chi(X_{t_1})\cdots\chi(X_{t_n})]$ is continuous; in particular, there exists a neighbourhood $\mathcal U$ of $x$ such that
$$ \mathbb E_y[\chi(X_{t_1})\cdots\chi(X_{t_n})] \geq 1-\frac\varepsilon{2|F|_\infty}$$
for any $y\in\mathcal U$.
Now that we found our cutoff function, it is time to conclude. For any $y\in\mathcal U$, we have
$$ P^n_tF(y)
 = \mathbb E_y[F(X_{t_1},\cdots,X_{t_n})\chi(X_{t_1})\cdots\chi(X_{t_n})]
 + \mathbb E_y\big[F(X_{t_1},\cdots,X_{t_n})(1-\chi(X_{t_1})\cdots\chi(X_{t_n}))\big]
 = (1)_y + (2)_y\text.$$
By construction, $|(2)_y|$ is at most $\varepsilon/2$. Moreover, according to the previous statement, $(1)_y$ is continuous with respect to $y$, and
$$ |(1)_x- P^n_tF(x)| \leq |F|_\infty\mathbb E_x[1-\chi(X_{t_1})\cdots\chi(X_{t_n})]
\leq \frac\varepsilon3\text. $$
In particular, we have $|(1)_y-P^n_tF(x)|<\varepsilon/2$ for all $y$ in a neighbourhood $\mathcal V$ of $x$, which proves that for any $y\in\mathcal U\cap\mathcal V$,
$$|P^n_tF(y)-P^n_tF(x)|<\varepsilon\text.$$
This concludes the proof.
