Let $B$ be a set of positive real numbers with the property that
adding together any finite subset of elements from $B$ always gives a sum of $2$ or
less. Show $B$ must be finite or countable.
Can this proof be criticized?
Let $m_1,m_2,m_3…$ be $B$'s (respectively) greatest element, second greatest element, third greatest element etc(the list could also be finite, or even empty if $B$ does not have a greatest element). I claim that the set $B\setminus\left\{m_1,m_2,…\right\}$ is empty, i.e. $B$ is either countable or finite:
Let $s$ be the supremum of $C=B\setminus\left\{m_1,m_2,…\right\}\neq\left\{\ \right\}$ because although it has no greatest element, it is bounded (all elements of $B\supseteq C$ are less than $2$ in particular).
For an arbitrary $\epsilon_1>0$ there is a $c_1$ in $C$ such that $s-\epsilon_1<c_1$; and there is a $\epsilon_2$ such that $0<\epsilon_2<s-c_1$(because $s$ is strictly greater than every element of $C$), and for such $\epsilon_2$ there is a $c_2>s-\epsilon_2>c_1$ in $C$ etc.:
$s-\epsilon_1<c_1<s-\epsilon_2<c_2<s-\epsilon_3<c_3<…<s\ \ $ i.e. there are $c$'s in $C$ such that $c_1<c_2<…$
Let $n$ be an integer greater than $\frac{2}{c_1}$: the finite sum $c_1+…+c_n>nc_1$ is greater than $2$. Contradiction in $C=B\setminus\left\{m_1,m_2,…\right\}\neq\left\{\ \right\}$.
Best Answer
I can elaborate and criticize. I see now that your idea is to eliminate all the large elements of the set provided they are isolated. So, yes, I agree that you are not assuming all the elements of $B$ are isolated. Even so the proof is flawed, but not completely misguided. (The error is, perhaps, the example that you give in your comment, $B= [0,1]\cup \{2,3,4\}$, mislead your intuition about this.)
You start: "Let $m_1,m_2,m_3,\dots$ be $B$'s (respectively) greatest element, second greatest element, third greatest element etc(the list could also be finite, or even empty if B does not have a greatest element). I claim that the set $B\setminus \{m_1,m_2,.\dots\}$ is empty."
Suppose that the set $B$ consists of all these numbers:
$$1, \dots, 1+\frac{1}{n}, 1+\frac{1}{n-1}, \dots, 1+\frac{1}{3}, 1+\frac{1}{2}$$ together with all of these
$$\frac12, \dots, \frac12+\frac{1}{n}, \frac12+\frac{1}{n-1}, \dots, \frac12+\frac{1}{3} $$
etc. if you want more.
Your argument does not work because the point $s$ that you claim exists is not an accumulation point of the new set $C$, so you cannot find that many values of $c$ in your set $C$ close to $s$.
Specifically: your list from largest down would be $$1+\frac{1}{2}, 1+\frac{1}{3}, 1+\frac{1}{4}, \dots $$ and would not end. So you would be left with $C$ (which is $B$ minus your list) and that contains the number $1$ so $s=1$. There does not exist a number $c_1$ with $s -\epsilon_1 <c_1< \dots$, etc.
There are simpler ideas to prove this, but let's go with your idea. You just want a positive number that is an accumulation point of $B$. But any set $B$ has a countable number of isolated points only. If there is no positive accumulation point then the set is countable. I think that is your proof. Is it?
For a nontopological, simply arithmetic proof, write the sequence of sets $$B_n=\left\{b\in B: b>\frac1n\right\}$$ and show each is finite.