More or less correct. But that is called an informal proof, not formal. People who already know how to prove the theorem will understand your explanation, but people who do not will not.
A formal proof would go something like:
$\def\nn{\mathbb{N}}$
$\def\rr{\mathbb{R}}$
[Assume that $(a_n)_{n\in\nn}$ is a sequence of real numbers, since it was not stated in the question.]
Let $b_n = \sum_{i=0}^n$, for any $n \in \nn$.
Given any $n\in\nn$:
$b_{n+1} \ge b_n$ because ...
Therefore $(b_n)_{n\in\nn}$ is increasing.
If $(b_n)_{n\in\nn}$ is bounded above:
$(b_n)_{n\in\nn}$ has a limit by ... theorem.
Thus $\sum_{i=0}^\infty$ converges by definition of infinite series.
Contradiction.
Therefore $(b_n)_{n\in\nn}$ is not bounded above.
Given any $m \in \rr$:
Let $k \in \nn$ such that $b_k > m$ by definition of boundedness.
Then $b_n \ge b_k$ for any $n \in \nn_{\ge k}$ because $(b_n)_{n\in\nn}$ is increasing.
Thus $b_n > m$ for any $n \in \nn_{\ge k}$.
Therefore $(b_n)_{n\in\nn}$ converges to $\infty$ by definition of infinite limit.
Therefore ...