Proving worst-case running time is in $\Omega(n^2)$
The worst-case running time of an algorithm $A$ is a function $T(n)$ which accepts an input length and outputs the maximum running time of algorithm $A$ on an input of length $n$:

$$ T(n) = \max_{|x|=n} \mathrm{runtime}(A,x). $$

Therefore to prove that $T(n) \geq S(n)$, it suffices to show that for each $n$ there is an input $x_n$ of size $n$ such that $\mathrm{runtime}(A,x_n) \geq S(n)$. The fact that there exist other inputs on which $A$ performs better is completely immaterial, since the worst-case running time depends on the worst input.