Consider the standard brownian motion $(W_{t})$, it is known that $W$ is a homogenous Markov process and thus it makes sense to consider if $W$ has Feller property.
Let $f$ be a bounded continuous function, to show $W$ is Feller, we need to show the semi-groups $$p^{t}f(x_{1})-p^{t}f(x_{2})\longrightarrow 0\ \ \text{as}\ \ x_{1}\longrightarrow x_{2}.$$
By definition we have \begin{align*} p^{t}f(x_{1})-p^{t}f(x_{2})&=\mathbb{E}[f(W_{t})|W_{0}=x_{1}]-\mathbb{E}[f(W_{t})|W_{0}=x_{2}]\\ &=\int_{\mathbb{R}}\mathbb{P}(t, x_{1}, dy)f(y)+\int_{\mathbb{R}}\mathbb{P}(t, x_{2}, dy)f(y). \end{align*} But we know the formula of transition probability of Wiener process: $$\mathbb{P}(t, x, \Gamma)=\int_{\Gamma}\dfrac{1}{\sqrt{2\pi t}}e^{\frac{-(y-x)^{2}}{2t}}dy,$$ hence the density must be $$\mathbb{P}(t, x, dy)=\dfrac{1}{\sqrt{2\pi t}}e^{\frac{-(y-x)^{2}}{2t}}dy.$$
This implies that $$\int_{\mathbb{R}}\mathbb{P}(t, x, dy)f(y)=\int_{\mathbb{R}}\dfrac{1}{\sqrt{2\pi t}}e^{\frac{-(y-x)^{2}}{2t}}f(y)dy,$$ and a change of variable $\tilde{y}+x:=y$ gives us $$\int_{\mathbb{R}}\dfrac{1}{\sqrt{2\pi t}}e^{\frac{-(y-x)^{2}}{2t}}f(y)dy=\int_{\mathbb{R}}\dfrac{1}{\sqrt{2\pi t}}e^{\frac{-\tilde{y}^{2}}{2t}}f(\tilde{y}+x)=\mathbb{E}(f(W_{t}+x)).$$
Therefore, \begin{align*} p^{t}f(x_{1})-p^{t}f(x_{2})&=\mathbb{E}(f(W_{t}+x_{1}))-\mathbb{E}(f(W_{t}+x_{2}))\\ &=\mathbb{E}\Big(f(W_{t}+x_{1})-f(W_{t}+x_{2})\Big). \end{align*}
What should I do next? Can I directly conclude as $x_{1}\rightarrow x_{2}$, $$\mathbb{E}\Big(f(W_{t}+x_{1})-f(W_{t}+x_{2})\Big)\longrightarrow 0?$$
Thank you!
Since $f$ is bounded, we know that $f(W_{t}+x_{1})-f(W_{t}+x_{2})$ is also bounded. And since $f$ is continuous, as $x_{1}\rightarrow x_{2}$, we must have $f(W_{t}+x_{1})-f(W_{t}+x_{2})\rightarrow 0$.
It then follows from bounded convergence theorem that $$p^{t}f(x_{1})-p^{t}f(x_{2})\longrightarrow 0\ \text{as}\ x_{1}\rightarrow x_{2}.$$