There are two pretty simple proofs. Both rely on studying the action $T_* \colon \mathcal{M} \to \mathcal{M}$, where $\mathcal{M}$ is the space of Borel probability measures on $X$ and the action is given by $(T_* \mu)(E) := \mu(T^{-1}(E))$. A measure $\mu$ is $T$-invariant if and only if $T_* \mu = \mu$.

One proof is the one given by Michael Coffey in his answer: start with any measure $\mu$, not necessarily invariant, such as the $\delta$-measure sitting at an arbitrary point, and then consider the sequence of measures $\mu_n = \frac 1n \sum_{k=0}^{n-1} T^k_* \mu$. Because $\mathcal{M}$ is weak* compact, some subsequence $\mu_{n_j}$ converges to a measure $\nu\in \mathcal{M}$, and it's not hard to show that $\nu$ is invariant.

An alternate proof is to observe that $\mathcal{M}$ is a compact convex subset of the locally convex vector space $C(X)^* $, and that $T_* $ acts continuously on $\mathcal{M}$, whence by the Schauder-Tychonoff fixed point theorem it has a fixed point $\nu=T_* \nu$.