Let $d$ be a real-valued function defined on $[a,b]$, for $a \lt b$, with the following property:
$\forall \varepsilon \gt 0: \exists N \in \mathbb N \cup \{0\}: \displaystyle\Big\lvert\{x\in[a,b]: |d(x)| \gt \varepsilon\} \Big\rvert = N$
As part of a larger proof, I want to show that $\displaystyle \inf_P U(d,P) \geq 0$. I proceeded by contradiction and would like to know whether or not I properly handled this contradiction. The contradiction leads to the statement, "the number of points in the set $\left.d\right|_{[t_{j-1},t_j]}$ is finite", which I believe is nonsense. I do not really have any experience with topological arguments or arguments involving infinities so any confirmation is appreciated.
Clarification: the notation "$\left.d\right|_{[t_{j-1},t_j]}$" refers to the collection of ordered pairs when $d$'s domain is restricted to the set $[t_{j-1},t_j]$.
Additionally, the notation "$\displaystyle \inf_P U(d,P)$" means the infimum of the set of all possible upper sum values of the function $d$ across every partition $P$ of $[a,b]$.
Suppose, instead, that there exists a partition $P$ of $[a,b]$ such that $U(d,P) \lt 0$. Let $P=\{a,t_1,\cdots,t_{n-1},b\}$. Equivalently, we have that $\displaystyle \sum_{i=1}^n M_i (t_i-t_{i-1}) \lt 0$, where $M_i$ is the supremum of $d$ on the corresponding subinterval $[t_{i-1},t_i]$. Because for any $i: \in \{1,2,\cdots,n\}: t_i-t_{i-1} \gt 0$, there must be at least one $j$ such that $ M_j \lt 0$. Let us now look at the corresponding subinterval $[t_{j-1},t_j]$ and describe the behavior of $d$.
Because $M_j \lt 0$ is the supremum of $d$ on $[t_{j-1},t_j]$, we know that $\forall x \in [t_{j-1},t_j]: d(x) \leq M_j \quad (*_1)$. Now, consider the value $\frac{M_j}{2} \lt 0$. From $d$'s property, we know that are a finite number of $x \in [a,b]: |d(x)| \gt -\frac{M_j}{2}$. In particular, then, there must be a finite number of $x \in [t_{j-1},t_j]:|d(x)| \gt -\frac{M_j}{2}$. Because all $d(x)$ in this subinterval are negative, we have that there are a finite number of $x \in [t_{j-1},t_j]: d(x) \lt \frac{M_j}{2}$. This implies that there must be a finite number of $x \in [t_{j-1},t_{j}]: d(x) \leq M_j$. However, from $(*_1)$, we know that there are no points in $[t_{j-1},t_j]$ that are mapped through $d$ to values greater than $M_j$. This means that the number of elements in the set $\left.d\right|_{[t_{j-1},t_j]}$ is finite. Note, though, that because $[t_{j-1},t_i]$ is a non-singleton interval of $\mathbb R$, if $d$ is restricted to this domain, then the number of ordered pairs contained within this subset of $d$ should be uncountably infinite. We therefore conclude that no such partition exists, which implies that $\displaystyle \inf_P U(d,P) \geq 0$.