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, if $M$ is homeomorphic to a connected Lie group then $M$ always admits a topologically transitive continuous self-map.
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.
Best Answer
There is a bit of a problem: at the end of your proof you conclude $\phi(w) = \phi(x)$, but we probably wanted to show that $\phi(y) = \phi(x)$?
How about this? I'm going to clarify your writing a little bit, because the first sentence of your proof is very long:
If you aren't quite sure why a continuous function which is constant on a dense subset of its domain is constant on the entire domain, the argument uses similar ideas to those from your original proof which have gone unused in mine above:
Lemma. If $f : X \to Y$ is continuous and constant on a dense subset $D \subset X$, then $f$ is constant on all of $X$.
Proof. By hypothesis $f(D) = \{ y \}$ for some $y \in Y$. Since $f$ is continuous the preimage of every closed subset of $Y$ is closed, hence $C = f^{-1}(\{y\}) \subset X$ is closed. Now $C$ is closed and contains $D$, so also contains the closure of $D$. But $D$ is dense in $X$, so the closure of $D$ is $X$---therefore $C = X$, too. It follows that $f(X) = \{y\}$, as desired.