[proofplan] We average Dirac masses along one forward orbit. Compactness of the space of Borel probability measures gives a weak-star cluster point. The telescoping identity for continuous test functions then shows that every such cluster point is invariant under $T$. [/proofplan] [step:Build empirical measures along one orbit] Choose $x_0\in X$. For $N\geq1$, define the empirical Borel probability measure \begin{align*} \mu_N:=\frac1N\sum_{n=0}^{N-1}\delta_{T^n x_0}. \end{align*} Since $X$ is compact metrizable, the space $\mathcal{M}(X)$ of Borel probability measures on $X$, equipped with the weak-star topology, is compact. Hence there is a subsequence $(\mu_{N_j})_{j\geq1}$ and a Borel probability measure $\mu$ such that \begin{align*} \mu_{N_j}\longrightarrow \mu \end{align*} in the weak-star topology. [/step] [step:Show that the weak-star limit is invariant] Let $f:X\to\mathbb{R}$ be continuous. Since $X$ is compact, $f$ is bounded. For every $N\geq1$, \begin{align*} \int_X f\circ T\,d\mu_N-\int_X f\,d\mu_N &= \frac1N\sum_{n=0}^{N-1} f(T^{n+1}x_0) - \frac1N\sum_{n=0}^{N-1} f(T^n x_0)\\ &= \frac{f(T^N x_0)-f(x_0)}{N}. \end{align*} Therefore \begin{align*} \left|\int_X f\circ T\,d\mu_N-\int_X f\,d\mu_N\right| \leq \frac{2\|f\|_\infty}{N}\to0. \end{align*} Because $T$ is continuous, $f\circ T$ is continuous. Passing to the subsequence limit gives \begin{align*} \int_X f\circ T\,d\mu=\int_X f\,d\mu \end{align*} for every continuous $f$. [/step] [step:Convert test-function invariance into measure invariance] For Borel probability measures on a compact metrizable space, equality of integrals against all continuous real-valued functions implies equality of measures. The preceding step therefore gives \begin{align*} T_\#\mu=\mu. \end{align*} Thus at least one $T$-invariant Borel probability measure exists. [/step]