Ok guys i think proved it, i used the following preposition which is equivalent for a dynamical topological pair $(X,T)$ to be uniquely ergodic :
For every continuous function $f\in C(X)$
$$\frac{1}{n}\sum_{k=0}^{n-1}f(T^k(x)) \to \int fd\mu \tag 1$$
for every $x\in X$ where $\mu$ is the uniquely determined $T-$invariant Borel probability measure.
Now i can prove the the desired result as follows:
Suppose we have a set $F$ which closed, $T-$invariant $\bigl(T(F)\subseteq F\bigr)$ and $F \neq X$ then $X\setminus F$ is non empty and open. Let $x_1 \in X\setminus F$, then there is an $\epsilon>0$ such that
$$E=\hat{B}(x_1,\epsilon) = \{x\in X: d(x,x_1)\leq \epsilon\} \subseteq X\setminus F$$
From our assumptions we get that $\mu(E)>0 $ since $\hat{B}(x_1,\epsilon)$ contains the open ball $B(x_1,\epsilon).$
Now we are trying to find the crucial continuous function $f$ in order to use $(1)$.Since we have two closed and disjoint sets $F,E$ we define
$$f(x) = \frac{dist(x,F)}{dist(x,F)+dist(x,E)}$$
then we know that
$\alpha)$ $f$ is continuous
$\beta)$ $0\leq f \leq 1$
$\gamma$) $f|F =0$ and $f|E=1$
So if we combine these facts and $(1)$ we get that
$$\frac{1}{n}\sum_{k=0}^{n-1}f\bigl(T^k(x)\bigr) \to \int fd\mu=\int_{X\setminus F}f d\mu \geq \int_{E}fd\mu \geq \mu(E)>0 \tag 2$$
for every $x\in X$.
But now (2) tells us that $F$ must be the empty set since if we can find $x^*\in F$ since $T(F) \subseteq F$ we get that $T^k(x^*) \in F$ for every k. In other words
$$\frac{1}{n}\sum_{k=0}^{n-1}f\bigl(T^k(x^*)\bigr) =0 $$
for every $n \in \mathbb{N}$ which contradicts $(2)$ since it holds for every $x\in X$.
So, for every $F$ closed which is $T-$invariant $\bigl(T(F)\subseteq F\bigr)$ it must be either $F= \emptyset$ either $F=X$.So, $(X,T)$ must be minimal.
Is this correct? thanks again !
Here's a counter-example for what you are trying to prove. As your question hints at, this is a very simple example where unique ergodicity holds yet some nonempty open set has measure zero.
Take $X = \mathbb R \cup \{P\}$ to be the one-point compactification of the real line $\mathbb R$. Define $f : X \to X$ by the formula $$f(x) = \begin{cases}
P & \quad\text{if $x=P$} \\
x+1 & \quad\text{if $x \in \mathbb R$}
\end{cases}
$$
The only invariant probability measure for this example is the Dirac measure on the point $P$, so $f$ is uniquely ergodic, but $\mathbb R = X - \{P\}$ is a nonempty invariant open subset.
Now define $Y = (-1,\infty) \subset \mathbb R$, which is an open subset of $X$. Note that $t(P) = +\infty$, and perhaps this might be what you would mean by saying that $t(P)$ is "not well-defined". And note also that while $t(x)$ is indeed well-defined for $x \in \mathbb R$, and while $t(x)=0$ for $x > -1$, on the other hand for any natural number $n$ any value of $x \in (-\infty,-n]$ satisfies $t(x) \ge n$, and so $t(x)$ has no upper bound.
My guess as to what is going on here is that you need to simply add some kind of topological hypothesis to your problem, for example that $f$ is topologically minimal, meaning that there is no proper invariant closed subset.
Best Answer
A uniquely ergodic system whose unique invariant measure is of full support is minimal.
For a uniquely ergodic system that is not minimal, just take a contraction $x \mapsto x/2$ on $[0,1]$.