Infinite sum of non-negative terms is equal to the supremum of the set of all finite sums

Hint: Say $s=\sum_{n=0}^\infty a_n$ and $\displaystyle S=\left\{\sum_{n\in F}a_n:F\in\mathscr P_{finite}(\Bbb N)\right\}$, so you want to show $s=\sup(S)$.

From what you say about the inequality $s\ge\sup(S)$ it seems clear you have the right idea; it could be much better phrased.

To show $s\le \sup(S)$, note first that it's enough to show that $$\alpha<\sup(S)\quad(\forall {\alpha<s}).$$But saying something is less than the least upper bound says it is not an upper bound. So what you need to show is this:

If $\alpha<s$ there exists a finite $F\subset\Bbb N$ with $$\alpha<\sum_{n\in F}a_n.$$