You want to prove a tight $\Theta(n)$ bound as @vonbrand commented you could to show:
$$am+b \leq T(m)\leq cm+d\quad (1)$$
Now start by assuming $(1)$ holds for all $m < n$ and conclude:
$$T(n) = 2T(n/3+1)+n \geq 2(a(n/3+1)+b)+n \geq n(2/3a+1) + 2(a+b)$$
$$T(n) = 2T(n/3+1)+n \leq 2(c(n/3+1)+d)+n \leq n(2/3c+1) + 2(c+d)$$
Thus $an+b \leq T(n)\leq cn+d$. ...
The definition of $\lceil x \rceil$ is:
$\lceil x \rceil$ is the minimal integer $n$ such that $n \geq x$.
(The existence of such an integer makes the reals an Archimedean field.)
Let us assume, for the sake of contradiction, that $\lceil x \rceil \geq x + 1$. Then $\lceil x \rceil - 1 \geq x$. Since $\lceil x \rceil - 1$ is also an integer, this ...
Let's start with the issue of iteration. Suppose that a function $f$ satisfies
$$ f(n/b) \leq (c/a)f(n). $$
Then it also satisfies
$$ f(n/b^2) \leq (c/a)f(n/b) \leq (c/a)^2 f(n). $$
You can prove by induction that for all integer $t \geq 0$,
$$ f(n/b^t) \leq (c/a)^t f(n). $$
As for your second question, about assuming that $n$ is large enough: the proof is ...