A sufficient condition for minimality in a Topological Dynamic System Am trying to prove the following :

Let ($X,T$) a uniquely ergodic topological system.We suppose that the uniquely $T-$invariant measure $\mu$ satisfies the property :$\mu(U)>0$ for every non-empty open set $U\subseteq X$.Then show under these conditions that the pair $(X,T)$ is minimal.

Let me clarify some definitions:
1)Topological System is the pair $(X,T)$ where $X $ is a compact metric space and $T:X\to X$ is a continuous transformation.
2)Minimal topological System is the pair $(X,T)$ where there does not exist a closed set $\emptyset \neq F \subseteq X$ such that $F\neq X$ and $T(F)\subseteq F$.
I tried to approach this by showing that every $x \in X$ has dense orbit but i only showed that 
$$\mu\biggl(\bigl\{x \in X: \overline{\{T^n(x):n \in \mathbb{N}_0\}}=X\bigr\}\biggr) =1$$
and i dont know has to use the assumption that $\mu$ is uniquely determined by being $T-$invariant.
Do you have any ideas? Thanks in advance !
 A: 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 !
