My approach would be simply to solve the recurrence: it’s not hard to see and to prove by induction that $T(n)=T(0)+\sum_{k=1}^nk=T(0)+\frac{n(n+1)}2=\frac12n^2+\frac12n+T(0)\;,$ and one can then appeal to the standard result that a polynomial of degree $k$ is $O(n^k)$. If you don’t have that result available, let $C=\max\left\{\frac12,|T(0)|\right\}$, and note that
$|T(n)|=\left|\frac12n^2+\frac12n+T(0)\right|\le C(n^2+n+1)\le 3Cn^2$
for $n\ge 1$.
I don’t follow your approach. If I were to do something along those lines, I’d assume that $T(k)\le ck^2$ for $k. Then
$\begin{align*} T(n)&=T(n-1)+n\\ &\le c(n-1)^2+n\\ &=cn^2-2cn+1+n\\ &=cn^2-(2cn-n-1)\\ &=cn^2-\big((2c-1)n-1\big)\;. \end{align*}$
You want to be sure that $T(n)\le cn^2$, so you want to be sure that $(2c-1)n-1\ge 0$. This will be the case provided that $c\ge\frac12\left(\frac1n+1\right)\;.\tag{1}$ The righthand side of $(1)$ is a decreasing function of $n$, so its maximum value is $1$, when $n=1$. Thus, there are two constraints on $c$: it must be at least $1$ to ensure that $(2c-1)n-1\ge 0$ for all $n\ge 0$, and it must be big enough so that $T(1)\le c$. Taking $c=\max\{1,|T(1)|\}$ does the job.