Is the following proof correct or not?
Please tell me better proof.
Let $\{c_n\}$ be a sequence of positive numbers such that $\sum c_n$ converges.
Let $S \subset S' \subset \mathbb{N}$.
Prove that $$\sum_{i \in S} c_i \leq \sum_{i \in S'} c_i.$$
If $|S| < +\infty$ and $|S'| < +\infty$, then it is obvious that $$\sum_{i \in S} c_i \leq \sum_{i \in S'} c_i.$$
If $|S| < +\infty$ and $|S'| = +\infty$, then it is obvious that $$\sum_{i \in S} c_i < \sum_{i \in S'} c_i.$$
So, assume that $|S| = +\infty$ and $|S'| = +\infty$.
Let $\phi$ be a bijection from $\mathbb{N}$ to $S$.
Let $\phi'$ be a bijection from $\mathbb{N}$ to $S'$.
Let $b_n = \sum_{i=1}^n c_{\phi(i)}$.
Let $b'_n = \sum_{i=1}^n c_{\phi'(i)}$.
For any $k \in \mathbb{N}$, there exists $l(k) \in \mathbb{N}$ such that $b_k \leq b'_{l(k)}$.
And we can choose $l(k)$ such that $l(k) < l(k+1)$ holds for any $k \in \mathbb{N}$.
Proof:
There exist $i_1, \cdots, i_k$ such that $\phi'(i_1) = \phi(1), \cdots, \phi'(i_k) = \phi(k)$ since $S \subset S'$.
Let $l(k) \in \mathbb{N}$ such that $l(k) > \max \{i_1, \cdots, i_k\}$.
Then, $\{\phi(1), \cdots, \phi(k)\} \subset \{\phi'(1), \cdots, \phi'(l(k))\}$.
So $b_k \leq b'_{l(k)}$.
And obviously we can choose $l(k)$ such that $l(k) < l(k+1)$ holds for any $k \in \mathbb{N}$.
Then, $$\sum_{i \in S} c_i = \lim_{k \to \infty} b_k \leq \lim_{k \to \infty} b'_{l(k)} = \sum_{i \in S'} c_i.$$
Without loss of generality, assume $S'$ is infinite. Now, the series either tends to infinity or converges to a number, $x$. In the former case, the result is trivial. In the latter case, since $S'\subseteq \mathbb N,$ we use the inclusion $\phi:S'\to \mathbb N$, noting the fact that any rearrangement of the terms gives the same sum, to write $\sum^{\infty}_{i=1}c_{\phi(i)}=x.$
Now, setting $T=S'\setminus S$, we have
$x=\sum^{\infty} _{i=1}c_{\phi(i)}= \sum_{\phi(i)\in S}c_i+\sum_{\phi(i)\in T} c_i\Rightarrow \sum_{S}c_i=x-\sum_T c_i\le x$
where we have used once more the fact that a convergent series of positive terms may be rearranged without changing the sum.