Continuous points of $f(x)=\inf_{g\in G}d(x, \varphi(g, x))$

Question

Let $\varphi:G\times X\to X$ be a continuous action on compact metric space $(X, d)$. The point $p\in X$ is called an equicontinuous, if for every $\epsilon>0$, there is $\delta>0$ such that

\begin{equation} d(y, p)<\delta \Rightarrow d(\varphi(g, y), \varphi(g, p))<\epsilon, \forall g\in G. \end{equation} Define $f(x)= \inf_{g\in G}d(x, \varphi(g, x))$.

In the following we try to show that if $x=p$ is an equicontinuous point for $\varphi$, then $f$ is continuous at $x=p$, (but I don't know that it is true or not. Please hep me to know it.)

Let $\epsilon>0$ be given. Choose $0<\delta<\frac{\epsilon}{2}$ corresponding to $\frac{\epsilon}{2}$ by equicontinuity of $x=p$. Since

$$d(p, \varphi(g, p))\leq d(p, y) +d(y ,\varphi(g, y))+ d(\varphi(g, y), \varphi(g, p))$$ Hence if $d(y, p)<\delta$, then $d(p,\varphi(g, p))< \epsilon +d(y, \varphi(g, y))$ for all $g\in G$, this implies that $f(p)< \epsilon+ f(y)$. Similarly we show that $f(y)<\epsilon+ f(p)$. Hence if $d(y, p)<\delta$, then $|f(y)-f(p)|<\epsilon$ i.e. $f$ is continuous at $x=p$.

Answer 1

"Similarly we show that":

notice that one cannot interchange $p$ and $y$ ($\delta$ is chosen for $p$, not for $y$).

On the other hand, we do have $$d(p, \varphi(g, p))\geq d(y ,\varphi(g, y))-d(p, y) - d(\varphi(g, y), \varphi(g, p)).$$ This yields the continuity of $f$.