An element generates the module, iff it does not belong to a maximal non-trivial submodule.
Because your module is 2-dimensional (as a vector space), the non-trivial maximal submodules must be 1-dimensional, and hence need to be eigenspaces of $T$. So if you pick a vector $v$ that is not an eigenvector it will not belong to any proper submodule, and hence generates all of $\mathbb{C}^2$.
If your matrix were larger, say 3x3, and three distinct eigenvalues, then the non-trivial maximal submodules would be direct sums of two eigenspaces. If there are repeated eigenvalues, then it gets a bit more complicated.
To answer your last question. The simplest counterexample is $T$= zero matrix. For any vector $v\in\mathbb{C}^2$, the $\mathbb{C}[T]$-submodule generated by $v$ is simply $\mathbb{C}v$. Hence the module $\mathbb{C}^2$ is not cyclic in that case. The same holds, if we let $T$ be any scalar matrix. There are other examples, [edit:added] when the matrices are larger than 2x2 (see further down)[/edit], but they all involve repeated eigenvalues.
In your example case we can select $v=(0,1)^T$. Then $T\cdot v=(1,1)^T$. You see that these two vectors are linearly independent over $\mathbb{C}$, and therefore they form a vector space basis.
In 3x3 case you can let $ T=\pmatrix{1&1&0\cr 0&1&0\cr 0&0&1\cr}. $ There are infinitely many maximal submodules $M$. We have $M_\infty$ spanned by $(1,0,0)^T$ and $(0,0,1)^T$ and $M_z$ spanned by $(1,0,0)^T$ and $(0,1,z)^T$, for any complex number $z$. The vector $(z_1,z_2,z_3)$ belongs to $M_\infty$, if $z_2=0$ and to $M_{z_3/z_2}$, if $z_2\neq0$. So any vector belongs to a non-trivial maximal submodule. Hence the module $\mathbb{C}^3$ is not cyclic.
The method of constructing non-cyclic modules in this way is to select $T$ in such a way that it has several Jordan blocks belonging to the same eigenvalue. Equivalently, the resulting module is cyclic, iff the characteristic and minimal polynomials of $T$ are equal. This follows from the theory of invariant factors. The minimal polynomial is a factor of the characteristic polynomial and has the same set of zeros, so for the two polynomials not to be equal it is necessary (but not sufficient) for $T$ to have repeated eigenvalues.