In the section of Spivak's Calculus on the construction of the reals, < is defined: if alpha and beta are real numbers, then a < b means that a is contained in b (that is, every element of a is also an element of b) but a doesn't equal b.
Does this imply that all negative numbers are contained in any positive number?
I don't have Spivak's book, but I assume he is using the definition of reals as Dedekind cuts of rationals. So we have already defined the ordered set of rationals $\mathbb Q$, and a real number is defined as a nonempty proper subset of $\mathbb Q$ that is downward-closed and has no greatest element.
Intuitively, a given real number is represented by the set of rationals less than it. So, under this definition, if $a$ and $b$ are reals with $a<b,$ we have $a\subset b.$ So, in particular, yes, any negative real is a subset of every positive real.
However, nobody would ever talk about two reals being subsets of one another outside of this very specific context. This is an unimportant detail of our particular set theoretical definition of the reals we have chosen. There are other perfectly valid definitions of the reals (e.g. as equivalence classes of Cauchy sequences of rationals) for which it wouldn't be true that negative reals were subsets of positive reals. This is what I meant when I called it an 'implementation detail' in the comments. The important thing about the reals for the purposes of analysis is the complete ordered field structure. The precise sets we choose to as the reals and which reals are members of one another in a set theoretical sense doesn't matter.