Prove that $\sum_{i \in S} c_i \leq \sum_{i \in S'} c_i$ if $S \subset S' \subset \mathbb{N}$.

Is the following proof correct or not?

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
$$x=\sum^{\infty} _{i=1}c_{\phi(i)}= \sum_{\phi(i)\in S}c_i+\sum_{\phi(i)\in T} c_i\Rightarrow \sum_{S}c_i=x-\sum_T c_i\le x$$