If the space has a countable basis and for the measure $\mu$ any open set has positive measure we can have this result as proved below, in other case there may be counterexample.
Since $X$ is a metric space, we can find countable basis $\{U_j\}_{j=1}^{\infty}$. Not that if for $x\in X$, the orbit of $x$ is not dense, then there exists some $U_j$ such that $\{T^n(x):n\in \mathbb{Z}\}\cap U_j=\emptyset$ so $$x\in A_j=X-\cup_{i=0}^{\infty}T^{-i}U_j.$$ So the set of $X$ such that the orbit of $x$ is not dense is contained in the set $\cup_{j=1}^{\infty} A_j$. Note that $A_j$ is $T$-invariant and has a empty intersection with $U_j$ so $\mu(A_j)=0$ since $U_j$ has positive measure. And so $\mu(\cup_{j=1}^{\infty} A_j)=0$ and the proof is completed.
Here is the answer rewritten to avoid the transfinite induction.
Given a topological space $X$, let $X'$ denote the set of non-isolated points of $X$.
Definition. Given a homeomorphism $T: X\to X$, a point $x\in X$ is called recurrent (with respect to $(X,T)$) if
for each neighborhood $U$ of $x$ there exists $n\ge 1$ such that $T^n(x)\in U$.
This condition is weaker than almost periodic.
Lemma 1. Let $X$ be a countable compact metrizable space, $T: X\to X$ a homeomorphism. Then every recurrent point
$x\in X$ is periodic.
Proof. Consider the collection ${\mathcal I}_x$ of all compact $T$-invariant subsets of $X$ containing $x$. Let $C_x$ denote the intersection of all members of ${\mathcal I}_x$. Clearly, $C_x\in {\mathcal I}_x$. I claim that $x$ is an isolated point
of $C_x$. Indeed, since $C_x$ is countable and compact metrizable, it has some isolated points, $C'_x\ne C_x$. If $x\in C'_x$ then $C'_x\in {\mathcal I}_x$ and $C'_x$ is strictly smaller than $C_x$, which is a contradiction. Hence, $x$ is isolated in $C_x$. The point $x$ is still recurrent with respect to $(C_x,T)$. Since $x$ is isolated in $C_x$, $\{x\}$ is a neighborhood of $x$ in $C_x$. Hence, by recurrence, there exists $n\ge 1$ such that $T^n(x)\in \{x\}$, i.e. $T^n(x)=x$, i.e. $x$ is $T$-periodic. qed
This answers Question 1. To answer Question 2, I will prove a stronger result:
Lemma 2. Let $X$ be a nonempty compact metrizable topological space, $T: X\to X$ is a homeomorphism. Then $X$ contains recurrent points. Equivalently, $X$ contains a $T$-invariant compact nonempty subset $X_0$ such that every $T$-orbit in $X_0$ is dense in $X_0$.
Proof. Consider the poset ${\mathcal I}$ of all nonempty $T$-invariant compact subsets of $X$ (with the partial order given by inclusion). Clearly, the intersection of members of each totally ordered (nonempty) subset in ${\mathcal I}$ belongs to ${\mathcal I}$. Hence, by Zorn's Lemma, ${\mathcal I}$ contains a minimal element $X_0$. By the minimality, every $T$-orbit in $X_0$ is dense (otherwise, take the closure of a non-dense $T$-orbit in $X_0$). qed
Combining the two lemmata, we see that if $X$ is countable, compact, metrizable, nonempty, then for each homeomorphism $T: X\to X$, there exists a $T$-periodic point.
Best Answer
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 :
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 !