Spivak, Ch. 3 Functions, Problem 28c: what exactly is being asked be proven? My issue with this item of this question is that it is not clear to me exactly what is being asked to be proven.

Let $F$ be the set of all functions whose domain is $\mathbb{R}$. Show
that $P10-P12$ cannot hold. In other words, show that there is no
collection $P$ of functions in $F$ such that $P10-P12$ hold for $P$.
(It will be sufficient, and will simplify things, to consider only
functions which are zero except a two points $x_0$ and $x_1$).

Note in what follows that it is pretty confusing already that $P$ is used to denote the collection of functions in $F$, given that $P$ denotes the collection of all positive numbers, and possibly all functions that are always positive (see below).
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$P10-P12$ are properties of real numbers defined in the first chapter. Here they are:

$P10$ (Trichotomy Law) For every number $a$, one and only one of the
following holds:
i) $a=0$
ii) $a$ is in the collection $P$
iii) $-a$
is in the collection $P$
where $P$ is the collection of all positive numbers.
$P11$ (Closure under addition) If $a$ and $b$ are in $P$, then $a+b$
is in $P$.
$P12$ (Closure under multiplication) If $a$ and $b$ are in $P$, then
$a\cdot b$ is in $P$.

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This is how I translate these properties if they were written for functions

$P10$ (Trichotomy Law) For every function $f$, one and only one of the
following holds:
i) $f(x)=0\ \forall x$
ii) $f$ is in the collection $P$
iii) $-f$ is in the collection $P$
where $P$ is the collection of all functions $f$ such that $f(x)>0\ \forall x$
$P11$ (Closure under addition) If functions $f_1$ and $f_2$ are in $P$, then $f_1+f_2$
is in $P$.
$P12$ (Closure under multiplication) If functions $f_1$ and $f_2$ are in $P$, then
$f_1 \cdot f_2$ is in $P$.

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The question as it is written seems to initially imply that we must show that $P10-P12$ cannot all hold for all functions in $F$. Then it says "in other words, show there is no collection $P$ of functions in $F$ such that $P10-P12$ hold for $P$".
I can find a subset of functions in $F$ that has all three properties:
Consider functions $f$ and $g$ such that $f(x)>0\ \forall x$ and $g(x)>0\ \forall x$.
Regarding $P10$, neither function is the constant zero function, and both are in $P$, but $-f$ and $-g$ are not in $P$. So $P10$ holds for both functions.
$f(x)+g(x) > 0\ \forall x$ so $P11$ holds.
$f(x)\cdot g(x) >0\ \forall x$ so $P12$ holds
Is the collection of functions with the two elements $f$ and $g$ not a collection such that all three of $P10$, $P11$, and $P12$ hold? Yet of course this doesn't mean all functions with real domain have all three properties.
Here is the solution in the solution manual

Let $f$ and $g$ be the two functions which are $0$ except at $x_0$ and
$x_1$, with $f(x_0)=1$ and $g(x_0)=0$, $g(x_1)=1$. Neither is $0$, so
$f$ or $-f$ would have to be in $P$, and likewise $g$ or $-g$. But
$(\pm f)(\pm g)=0$, which contradicts $P12$.

 A: Your collection $P = \{ f, g \}$ of two functions satisfies none of the three properties:
P10 is not satisfied because in it, you need to consider all functions, not just those in $P$. Both of your functions are positive everywhere. So if I give you a function $h$ that is positive at one point and negative at another, neither $h$ nor $-h$ will be among the two functions in your set. It it also not the zero function. (This also shows that it won’t help to enlarge your set to include all functions that a positive everywhere.)
P11 is not satisfied, because you don’t need to show that the function $f + g$ defined by
$$
  (f + g)(x) = f(x) + g(x)
$$
is positive everywhere; you need to show that it is in your set $P$. But for any $x$, $f(x) + g(x)$ will be bigger than both $f(x)$ and $g(x)$, so it can be neither of the functions $f$ or $g$.
P12 might be satisfied in very particular circumstances (namely if both $f$ and $g$ are $1$ everywhere); in all other cases, you run into similar problems as for P11.

Let me try to clarify what you are tasked to do. The exercise wants you to show that it is impossible to find a subset $P$ of the set of all functions $F$ with the properties P10, P11, P12. (You should ignore that the book might have defined a set $P$ of numbers earlier, there just aren’t enough letters in the alphabet to use a new one every time.)
Where previously, the properties talked about numbers, you now need to translate them to use functions. You did this correctly, except that you shouldn’t have the clause “where $P$ is the collections of all functions $f$ such that $f(x) > 0~\forall x$” in there, because this would define $P$ (albeit not with the required properties). Instead, you need to show that this is impossible, no matter how you choose $P$. For example, it also does not work if choose the set of all functions such that $f(0) > 0$.
This is what the solution from the solution manual does. It assumes that you are given a set $P$ with the three properties P10, P11, P12. Then, it constructs two functions $f$ and $g$ where $f$ is zero everywhere except at $x_0$ and $g$ is zero everywhere except at the point $x_1$ (with $x_0 \neq x_1$). Now, it only uses the properties P10, P11, P12. Because of P10, either $f \in P$ or $-f \in P$ and either $g \in P$ or $-g \in P$. (For both functions, we know that the case (i) in P10 is not the case, so (ii) or (iii) must be true if $P$ has the property P10.) Let’s first consider the case that $f, g \in P$. The product function $f g$ is zero everywhere, so it can’t be in $P$ by P10. But by property P12 it needs to in $P$; so we have a contradiction. The other cases that $-f, g \in P$, $f, -g \in P$ or $-f, -g \in P$ are similar. The solution in the manual shortens this to $(\pm f)(\pm g) = 0$.
