Feller continuity of the stochastic kernel

No. Take $X = \mathbb{R}$ and $K(x,B) = 1_B(x+1)$. (This corresponds to a process that jumps 1 unit to the right at each step.) Note that $\int f(y) K(x,dy) = f(x+1)$, so $K$ is certainly weak Feller.

However, take $f = 1_{(0, \infty)}$ to be a step function and $B = \mathbb{R} \backslash \{0\}$. Then $f \in C(B)$ by your notation. But I'll let you check that $\int_B f(y) K(x, dy) = 1_{(-1, \infty)}(x)$ which is not in $C(B)$.