Here is the definition of limit:
$\lim\limits_{x\rightarrow a} f(x)=L$ if for every $\epsilon>0$, there is a $\delta>0$ such that $|f(x)-L|<\epsilon\quad\text{whenever }\quad 0<|x-a|<\delta.$
What does it mean that $\lim\limits_{x\rightarrow a} f(x)\ne L$?
Well, it means (imprecisely) that there is an $\epsilon>0$ such that $f(x)$ is not close to $L$ no matter how close $x$ is to $a$.
More precisely it means that there is an $\epsilon>0$, such that no matter how small $\delta>0$ is, there is an $x$ with $0<|x-a|<\delta$ and yet $|f(x)-L|\ge\epsilon$.
So, in your case, you need to find a fixed value of $\epsilon$ such that for any $\delta>0$ there is an $x$ such that the following holds: $ \tag{1}|(x+3)-6|\ge\epsilon\quad\text{and}\quad 0<|x-2|<\delta. $
Here, you could choose $\epsilon=1/2$.
Given any $\delta>0$, choose any $x$ such that $0<|x-2|<\min\{\delta,1/2\}$.
Then $x$ would be in the interval $(1.5,2.5)\,$. Consequently, $x+3$ would be in the interval $(4.5,5.5)$ and thus at least $1/2$ units away from 6. That is, $|(x+3)-6|\ge1/2$.
Informally, if $x$ is very close to 2, then $x+3$ would be far away from 6. And so there would be no $\delta$ that "works" in the definition of limit. The quantity $x+3$ is at least 1/2 unit away from 6 whenever $x$ is within 1/2 of 2.