That's too ambitious, there is an enormous variety of metric spaces. But if you restrict to specific classes, then a lot has been done. I can think of two examples: Hilbert spaces and geodesic surfaces. These are completely classified.
Hilbert spaces are vector spaces equipped with a scalar product, which induces a norm, hence a distance and so a structure of metric space, which is required to be complete. Hilbert spaces are completely characterized up to isomorphism (a stronger form of isometry); in particular, any separable real Hilbert space is isomorphic to
$$
\ell^2:=\left\{\boldsymbol{x}=(x_1, x_2, x_3, \ldots)\ :\ x_j\in\mathbb R,\ \sum_{j=1}^\infty x_j^2<\infty\right\},$$
where the scalar product is given by
$$
\langle \boldsymbol{x}, \boldsymbol{y}\rangle = \sum_{j=1}^\infty x_j y_j.$$
(Non-separable Hilbert spaces are classified in terms of cardinality of their orthonormal bases. Also, the same result holds in the complex case, with obvious modifications).
Geodesic surfaces are something I know less. I will refer to this beautiful note of Etienne Ghys, section 3. It is in French, but I am sure I have seen an English translation on the net.
Suppose that $f$ is not transitive.
Then there are non-empty open subsets $U,V$ such that $f^n(U) \cap V = \emptyset$ forall $n \ge 0$.
Using your hypothesis with some open balls contained in $U$ and $V$, we can deduce that there is a trajectory that goes through $V$ and then through $U$ : there is a $w \in X$ and integers $0\le p<q$ such that $f^p(w) \in V$ and $f^q(w) \in U$.
Let $n=q-p \ge 0$. Since $f^n$ is continuous, $(f^n)^{-1}(U)$ is open, and letting $V' = V \cap (f^n)^{-1}(U)$ (a non-empty open set), we have $f^n(V') \subset U$.
Now, we know that any point in $V'$ will be sent to $U$ after $n$ steps. Since it is forbidden to visit $V$ after visiting $U$, this shows that a trajectory cannot visit $V'$ too many times.
If $m_0$ is the smallest index such that $f^{m_0}(w) \in V'$, then $f^{m_0+n}(w) \in U$ and so from $w \ge m_0+n$, $f^m(w) \notin V'$, which leaves at best $n$ indices $m_0,m_0+1,\ldots, m_0+n-1$ for which it is allowed to have $f^m(w) \in V'$.
Then, to reach a contradiction, we only have to show that forall $n$ there exist trajectories that visit $V'$ at least $n$ times.
We show this by induction. It is fairly obvious that there is a trajectory that goes through $V'$ at least once.
Now let $n \ge 1$ and suppose we have a trajectory that visits $V'$ at least $n$ times : we have a $w \in X$ and $n$ indices $m_1 \ldots m_n$ such that $w_i = f^{m_i}(w) \in V'$. If any two $w_i$ are equal then the trajectory is periodic and we are done. If not, since $w_1$ isn't an isolated point, there is a point $z \in V'$ such that $z$ is distinct from all the $w_i$.
Then pick $d>0$ such that the open balls $B_i = B(w_i,d)$ and $B_0 = B(z,d)$ are pairwise disjoint and contained in $V'$. Since all the $f^{m_i-m_1}$ are continuous, we can find $\epsilon > 0$ such that $f^{m_i-m_1}(B(w_1,\epsilon)) \subset B_i$ forall $i \in \{1 \ldots n\}$.
Now we apply your hypothesis with $\epsilon$, $z$, and $w_1$. This tells us that there exists a trajectory that comes within $\epsilon$ of $z$ and $w_1$. From our choice of $\epsilon$, we deduce that this trajectory comes within $d$ of $z$ and all of the $w_i$. Since those $n+1$ balls are disjoint and contained in $V'$, this trajectory does meet $V'$ at least $n+1$ different times.
Best Answer
No. As an example take $X$ which is the disjoint union of the Knaster–Kuratowski fan $X_1$ and interval $X_2=[0,1]$. Since $X_1$ contains a dispersion point, every continuous map $X_2\to X_1$ is constant. Hence, for every continuous map $T: X\to X$ either $T(X_1)\subset X_1, T(X_2)\subset X_2$ (in which case $T$ of course cannot be topologically transitive) or $T(X)\subset X_1$ (ditto), or $T(X)\subset X_2$ (ditto), or $T(X_1)\subset X_2$, $T(X_2)=\{x_1\}\subset X_1$. Consider the last case and set $x_2=T(x_1)\in X_2$. Then for every $n\ge 2$, $T^n(X)\subset \{x_1, x_2\}$. It is clear that such $T$ cannot be topologically transitive either.
Edit 1: Actually, one can even take $X=(-1,0) \cup [1,2]$. A similar argument will work since (apart form three trivial cases) the image of $[1,2]$ in $(-1,0)$ will be a compact subinterval $I$ and $T^n(X)\subset I \cup [1,2]$ for all $n\ge 2$, hence, $T$ cannot be topologically transitive.
Edit 2: It follows from the results of
S. Alpern and V. S. Prasad: Typical dynamics of volume preserving homeomorphisms. Cambridge Tracts in Mathematics, 139, Cambridge University Press, Cambridge, 2000.
that for every connected compact or open and tame manifold $M$ of dimension $\ge 2$, there exists a topologically transitive homeomorphism $T: M\to M$.
Here a manifold $M$ is called tame if it is homeomorphic to the interior of a compact manifold with boundary. In particular, each ${\mathbb R}^n$, $n\ge 2$, admits a topologically transitive self-homeomorphisms.
Remark. It appears that the first example of such self-map of the plane is due to L.Shnirelman, 1930 (but published in a barely accessible place). The most common reference for topologically transitive homeomorphisms of the plane is
A. S. Besicovitch, A problem on topological transformation of the plane, Fund. Math. 28(1937), 61-65.
As for connected manifolds $M$ dimension 1, while (apart from circle) there will be no topologically transitive homeomorphisms. However, there is always a topologically transitive continuous self-map. For instance, if $M={\mathbb R}$, then an example is given by
$$ T(x)= \begin{cases} −3x+8n+2, & \hbox{if} ~~2n\le x\le 2n+1\\ 5x−8n+2, & \hbox{if} ~~2n−1\le x\le 2n \end{cases} $$ with $n\in {\mathbb Z}$. See
A. Nagar and S. P. Sesha Sai, Some classes of transitive maps on ${\mathbb R}$, Jour. of Anal., 8 (2000), 103–111.
In particular, each finite-dimensional Banach space admits a topologically transitive continuous self-map. This leaves open the case of infinite-dimensional separable Banach spaces. (As Dan Rust noted in a comment, there is only one such space up to a homeomorphism, so it suffices to understand the case of $\ell_2$.)
As for compact spaces, the situation is unclear. It is known that if $M$ is a compact metrizable finite-dimensional Peano continuum (i.e. a connected and locally connected space) then $M$ admits a topologically transitive continuous self-map:
S. Agronsky and J. G. Ceder: Each Peano subspace of $E^k$ is an $\omega$-limit set, Real Anal. Exchange 17 (1991/92), no. 1, 371-378.
This gives yet another proof that each compact connected manifold (possibly with boundary) admits a topologically transitive continuous self-map.