Here is what I got for a compact metric space:
We must show that any sequence $\mu_n$ in $\mathscr{M}(X)$ has a $\omega^*$-convergent subsequence (this shows that $\mathscr{M}(X)$ is actually sequentially compact).
Let $\{f_i\}_{i=1}^{\infty}$ be a countable dense subset of functions in $C(X)$. For any sequence $\mu_n$ in $\mathscr{M}(X)$ we have that \begin{equation*} |\int_X f_1 d\mu_n| \leq \|f_1\|_\infty \end{equation*} for all $n$, hence the sequence $\int_X f_1 d\mu_n$ is bounded and therefore has a convergent subsequence which we will denote by $\int_X f_1 d\mu_{n}^{1}$.
Now consider the sequence $\int_X f_2 d\mu_{n}^{1}$. It is again a bounded sequence of real numbers and so it has a convergent subsequence $\int_X f_2 d\mu_{n}^{2}$.
We continue in this fashion and obtain, for each $i \geq 1$, \begin{equation*} ... \subset \mu_{n}^{i} \subset \mu_{n}^{i-1} \subset ... \subset \mu_{n}^{1} \end{equation*} such that $\int_X f_j d\mu_{n}^{i}$ converges for $1 \leq j \leq i$. Now consider the sequence $\mu_{n}^{n}$. Since, for $n \geq i$, $\mu_{n}^{n}$ is a subsequence of $\mu_{n}^{i}$, $\int_X f_i d\mu_{n}^{n}$ converges for every $i \geq 1$. \
We can now use the fact that $\{f_i\}_{i=1}^{\infty}$ is dense to show that $\int_X f d\mu_n$ converges for all $f \in C(X)$. For any $\epsilon > 0$, choose $f_i$ such that $\|f-f_i\|_\infty \leq \epsilon$. Since $\int_X f_i d\mu_{n}^{n}$ converges, there exists $N$ such that if $n,m \geq N$ then \begin{equation*} \lvert \int_X f_i d\mu_{n}^{n} - \int_X f_i d\mu_{m}^{m} \rvert < \epsilon. \end{equation*}
Thus if $n, m \geq N$ we have \begin{equation*} | \int_X f d\mu_{n}^{n} - \int_X f d\mu_{m}^{m} | \leq \end{equation*} \begin{equation*}| \int_X f d\mu_{n}^{n} - \int_X f_i d\mu_{n}^{n} | + | \int_X f_i d\mu_{n}^{n} - \int_X f_i d\mu_{m}^{m} | + | \int_X f_i d\mu_{m}^{m} - \int_X f d\mu_{m}^{m} | \leq 3 \epsilon. \end{equation*}
So, $\int_X fd\mu_{n}^{n}$ converges, as required.
To complete the proof, write \begin{equation*} \Lambda (f) = \lim_{n \rightarrow \infty} \int_X f d\mu_{n}^{n}. \end{equation*} It is easily verified that $\Lambda$ satisfies the hypothesis of the Riesz Representation Theorem. Therefore, there exists $\mu \in \mathscr{M}(X)$ such that \begin{equation*} \Lambda (f) = \int_X f d\mu.\end{equation*} We then have \begin{equation*} \int_X f d\mu_{n}^{n} \rightarrow \int_X f d\mu, \end{equation*} as $n \rightarrow \infty$, for all $f \in C(X)$, so we have proved that $\mu_{n}^{n}$ $\omega^*$-converges to $\mu$, as $n \rightarrow \infty$ and we are done. $\blacksquare$
After thinking on it, I think that these different ergodic measures will necessarily be singular. Basically, if $\mu, \nu$ are distinct ergodic measures, then there's some Borel $A \subseteq I$ such that $\mu(A) = \nu(A)$. Then we know that
\begin{align*}
\mu \left( \left\{ x \in I : \lim_{k \to \infty} \frac{1}{k} \sum_{j = 1}^k \chi_A \left( T^{j - 1}(x) \right) = \mu(A) \right\} \right) & = 1, \\
\nu \left( \left\{ x \in I : \lim_{k \to \infty} \frac{1}{k} \sum_{j = 1}^k \chi_A \left( T^{j - 1}(x) \right) = \nu(A) \right\} \right) & = 1 .
\end{align*}
But these two sets must necessarily be disjoint.
Best Answer
For any measurable space $(\Omega, \mathcal F)$, the natural sigma-algebra $\mathcal D$ on the set $\Delta$ of probability measures on $(\Omega, \mathcal F)$ is defined as follows. For each $A \in \mathcal F$, let $X_A: \Delta \to \mathbb R$ be the function defined by $X_A(P) = P(A)$. Define $\mathcal D$ to be the smallest sigma-algebra on $\Delta$ that makes every member of $\{X_A: A \in \mathcal F\}$ measurable (we assume, as is standard, that $\mathbb R$ is equipped with its Borel sigma-algebra). Equivalently, $\mathcal D$ is the sigma-algebra generated by $\{X_A^{-1}(B): A \in \mathcal F, B \subseteq \mathbb R \ \text{Borel}\}$.