Show that $X$ is countable. 
Let $X \subset \mathbb{R}_{\geq 0 }$. Suppose there exists $C>0$ such that for any finite subset $\{x_1,x_2,...,x_n\}\subset X$ $\sum_{i=1}^{n}x_i \leq C$. Show that $X$ is countable.

I am quite lost in trying to solve this exercise. I've only thought about $\mathcal{P}_{<\infty}(X)=\{A\subset X:\#(A)<\infty\}$, and I found out that $\#(\mathcal{P}_{<\infty}(X))=\aleph_{0}$ if $\#(X)=\aleph_{0}$ and $(\mathcal{P}_{<\infty}(X))=\mathfrak{c}$ if  $\#(X)=\mathfrak{c}$. However I can't go any further, and I don't even know if this is useful somehow. Any hint?
 A: Consider
$$
X_n=X\cap(1/n,\infty).
$$
Then $\bigcup_{n=1}^\infty X_n=X$. If all the $X_n$'s were countable, then so would be $X$. Hence, for some $m\in\mathbb N$, we have that $\lvert X_{m}\rvert>\aleph_0$. Let $x_1,\ldots,x_N\in X_m$, where $N>cm$.
Then
$$
x_1+\cdots+x_N>\frac{N}{m}>c.
$$
A: Assume some point of $X$ is an interior point, $x$. Fix $r>0$ such the neighborhood of radius $r$ around $x$ is contained in $X$. This neighborhood has infinitely many points of $X$.
Choose $k$ such that $kx>C$, then pick $k-1$ many $x_i$ from the neighborhood around $x$ such that each $x_i>x$. Then the sum of all $k$ elements is greater than $C$, thus by contradiction, no point of $X$ can be an interior point.
Then any point in $X$ must be either an isolated point or a limit point of $X$ (or both). The set of isolated points is obviously countable, so the question becomes how many limit points there are.
For any nonzero limit point, there are infinitely many points in $X$ in any neighborhood of that limit point. Thus, by a similar process as above, we can create another contradiction to show that no such limit point exists. Of course, a limit point at $0$ is still possible, but that is not a problem in terms of countability.
