Let us also note that (some) separability is also needed for the existence of a point with dense orbit (by definition).
As an explicit example, consider the space $X_0=F(A;[0,1])$ of all functions from a set $A$ to $[0,1]$, where $A$ is with $\operatorname{card}(A)>\operatorname{card}(\mathbb{R})$, the target is endowed with discrete topology and $X_0$ is endowed with product topology (= topology of pointwise convergence). Then $X_0$ is compact Hausdorff non-separable (see https://math.stackexchange.com/a/3360831/169085). Then define the space $X=F(\mathbb{Z};X_0)$ of bi-infinite sequences in $X_0$, again with product topology. Then $X$ too is compact Hausdorff (see The product of Hausdorff spaces is Hausdorff) and non-separable. Define $T:X\to X$ to be the shift homeomorphism, $\omega_\bullet\mapsto \omega_{\bullet+1}$. Then $T$ is topologically strong mixing (as defined in Topological weakly mixing implies totally transitive), hence also topologically transitive.
In fact, analyzing the proof of the Birkhoff Transitivity Theorem one can obtain an optimal statement. For a continuous self-map $T:X\to X$ of a topological space $X$, for a point $x\in X$ and two subsets $A,B\subseteq X$ put
$$\mathcal{O}_x(T)=\{T^n(x)\,|\, n\in\mathbb{Z}_{\geq0}\},$$
$$\mathcal{D}(T)=\{x\in X\,|\, \overline{\mathcal{O}_x(T)}=X\},$$
$$N^T(A\to B)=\{n\in\mathbb{Z}_{\geq0}\,|\, T^n(A)\cap B\neq\emptyset\},$$
so that $\mathcal{O}_x(T)$ is the orbit of $x$ under $T$, $\mathcal{D}(T)$ is the subset of $X$ that consists of all points whose $T$-orbits are dense, and $N^T(A\to B)$ is the set of return times from $A$ to $B$. Denote by $\mathcal{T}(X)^\ast$ the collection of all nonempty open subsets of $X$. $T$ is topologically transitive if
$$\forall U,V\in\mathcal{T}(X)^\ast: N^T(U\to V)\neq\emptyset.$$
Theorem: Let $X$ be a Hausdorff ($\ast$) topological space, $T:X\to X$ be a continuous self-map. Then
- If $\mathcal{D}(T)$ contains a non-isolated point ($\dagger$), then $T$ is topologically transitive.
- If $T$ is topologically transitive, and $X$ is second countable (= its topology has a countable base) ($\dagger\dagger$) and Baire (= intersection of countably many open dense subsets is dense), then $\mathcal{D}(T)$ is a dense $G_\delta$.
Proof: (1) Let $x^\ast\in \mathcal{D}(T)$ be non-isolated. Then any open neighborhood of $x^\ast$ must contain points other than $x^\ast$. Let $U\in\mathcal{T}(X)^\ast$. We claim that $N^T(\{x^\ast\}\to U)$ is infinite. Indeed, by the density of the orbit of $x^\ast$, there is an $n_1\in N^T(\{x^\ast\}\to U)$; whence $T^{-n_1}(U)$ is an open neighborhood of $x^\ast$. Since $x^\ast$ is not isolated (and singletons are closed) $T^{-n_1}(U)\setminus\{x^\ast\}$ is also nonempty open. Using the density again there is an $m_1\in\mathbb{Z}_{\geq0}$ such that $T^{m_1}(x^\ast)\in T^{-n_1}(U)\setminus\{x^\ast\}$; thus by construction $T^{n_2}(x^\ast)\in U$ for $n_2=n_1+m_1>n_1\geq 0$, i.e. $n_2\in N^T(\{x^\ast\}\to U)\cap \mathbb{Z}_{\geq1}$. By induction $N^T(\{x^\ast\}\to U)$ is infinite; that is, the orbit of $x^\ast$ hits $U$ infinitely many times.
Now let $U,V\in\mathcal{T}(X)^\ast$ and pick $a\in N^T(\{x^\ast\}\to U)$ and $b\in N^T(\{x^\ast\}\to V)\cap \mathbb{Z}_{>a}$ (such an $a$ exists by assumption and such a $b$ exists because $N^T(\{x^\ast\}\to V)$ is infinite). Then $b-a\in N^T(U\to V)\cap \mathbb{Z}_{\geq1}$.
(2) Let $U_\bullet:\mathbb{Z}_{\geq1}\to\mathcal{T}(X)^\ast$ be an ordered base for the topology of $X$. Then we have:
$$\mathcal{D}(T)=\bigcap_{m\in\mathbb{Z}_{\geq1}}\bigcup_{n\in\mathbb{Z}_{\geq0}} T^{-n}(U_m).$$
By topological transitivity, $\bigcup_{n\in\mathbb{Z}_{\geq0}} T^{-n}(U_m)$ is dense for any $m\in\mathbb{Z}_{\geq1}$; any such set is also open. Since $X$ is Baire, $\mathcal{D}(T)$ is dense; it's also $G_\delta$ by construction.
($\ast$) Singletons being closed is enough, i.e. $T_1$.
($\dagger$) This is needed, e.g. consider $T:\mathbb{Z}_{\geq0}\to \mathbb{Z}_{\geq0}, x\mapsto x+1$; which is not topologically transitive (one can't go back) but $\mathcal{D}(T)=\{0\}\neq\emptyset$.
($\dagger\dagger$) Note that this is a priori stronger than separability.
Finally, to compare, note that in FShrike's example $X=[0,1]\cap\mathbb{Q}$ is not a Baire space (let alone completeness; also see Example of a Baire metric space which is not completely metrizable).
