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 ...
Well, you have almost everything you need. Let $x\in[0,\infty)$ and let $\epsilon>0$. We want to show that there is a finite subsequence of $(a_n)$ such that the sum of its elements is in the interval $(x-\epsilon, x+\epsilon)$. As you already know $x$ is the sum of all elements in a subsequence $(a_{n_k})$ of $a_n$. If the subsequence is finite then we are done because $\sum a_{n_k}=x\in(x-\epsilon, x+\epsilon)$ and that's what we need. Now suppose the subsequence is infinite. We still know that $\sum_{k=1}^\infty a_{n_k}=x$. An equivalent way to write it is $\lim_{M\to\infty}\sum_{k=1}^M a_{n_k}=x$. By the definition of the limit we can pick a large enough $M$ such that $\sum_{k=1}^M a_{n_k}\in(x-\epsilon,x+\epsilon)$. And $(a_{n_k})_{k=1}^M$ is a finite subsequence of $(a_n)$.
Best Answer
Let the interval be $(x,y)$.
Let $b_i$ be the subsequence of the $a_i$ with $ a_i < y-x$.
$b_i$ excludes at most a finite number of $a_i$ and therefore its sums also diverge.
Now let $n$ be the largest value with $\sum\limits_{i=1}^n b_i \leq x$ and note that $x<\sum\limits_{i=1}^{n+1}b_i < x + (y-x) = y$.