If I form pairs for every value in my set: $(0, \ldots, n)$, without replacement: (ie, I form $0,1$ but not $1,0$), how many pairs can I form, in terms of $n$, with there still being the possibility that one of the values from $0$ to $n$ has not yet been used in a pair?
Is this $(n-1)+(n-2)+(n-3)....+1$?
Your answer is correct.
The set $\{0, 1, 2, 3, \ldots, n\}$ has $n + 1$ elements, so there are $\binom{n + 1}{2}$ two-element subsets. Select a particular element, say $0$. Since there are $n$ other elements in the set, there are $\binom{n}{2}$ ways to select two-element subsets that do not include $0$. Thus, the largest number of two-element subsets that can be selected from $\{0, 1, 2, \ldots, n\}$ without using a particular element is $$\binom{n}{2} = \frac{n(n - 1)}{2} = \sum_{k = 0}^{n} k = 0 + 1 + 2 + \cdots + (n - 3) + (n - 2) + (n - 1)$$ as you found.
To understand why $$\binom{n}{2} = \sum_{k = 0}^{n - 1} k$$ consider two-element subsets of $\{1, 2, 3, \ldots, n\}$ whose largest element is $j$. In each case, there are $j - 1$ ways to choose the smaller element of the subset, so the number of two-element subsets is $$\binom{n}{2} = \sum_{j = 1}^{n} (j - 1) = \sum_{k = 0}^{n - 1} k$$ where $k = j - 1$.
Thus, to ensure that every element appears in at least one two-element subset of $\{0, 1, 2, \ldots, n\}$, we must select $$\binom{n}{2} + 1$$ subsets.