There is a real number $\xi$ such that one can neither prove nor disprove that $\xi$ is positive 
There is a real number $\xi$ such that one can neither prove nor disprove that $\xi$ is positive.
Proof: There are uncountably many real numbers but only countably many proofs.

This is something I just thought of. I don't know anything about proof theory or anything like that. Is my reasoning correct?
Just to emphasize, I would like to know if the proof is correct. The statement itself is true, see for example Yves Daoust's answer.
 A: Your question is ill-defined, simply because proofs are strings of symbols, and how will you describe every real number by a string of symbols, so that it can even be part of a proof? Based on that alone you cannot talk about proving or disproving some sentence about an arbitrary real number. If that answers your inquiry, then that is all there is to it.
$
\def\rr{\mathbb{R}}
$
That said, there is an interesting separate question that can be made out of it:

Is there a real number $x$ that is definable over ZFC such that ZFC cannot prove or disprove that $x > 0$?

The above statement can easily be made precise as follows:

Let $\rr$ be the set of real numbers as defined in ZFC, and $0_\rr$ be the zero element in $\rr$. Is there a $1$-parameter sentence $P$ over ZFC such that ZFC proves "$\exists! x ( P(x) \land x \in \rr )$" but ZFC proves neither "$\forall x ( P(x) \to x > 0_\rr )$" nor "$\forall x ( P(x) \to x \le 0_\rr )$".

Notice that this question cannot be easily answered by resorting to cardinality considerations. But the answer to this question is, curiously, "yes". Let $1_\rr$ be the unit element in $\rr$. (Actually all that matters is that ZFC proves "$1_\rr > 0_\rr$".) Let $Q$ be some independent sentence over ZFC, such as $\neg \text{Con}(\text{ZFC})$, and let $P(x) \overset{def}\equiv Q \land x=1_\rr \lor \neg Q \land x=0_\rr$. Then ZFC proves "$P(0) \lor P(1)$" and "$\forall x,y ( P(x) \land P(y) \to x=y )$", and hence also proves "$\exists! x ( P(x) \land x\in\rr )$". However, ZFC cannot prove "$\forall x ( P(x) \to x > 0_\rr )$" otherwise it would also prove "$\neg P(0_\rr)$" and hence "$Q$". Similarly ZFC cannot prove "$\forall x ( P(x) \to x \le 0_\rr )$" otherwise it would also prove "$\neg P(1_\rr)$" and hence "$\neg Q$".
So you can see from this that the syntactic incompleteness of ZFC has an effect on the decidability of even basic questions about definable reals.
A: 
Proof: There are uncountably many real numbers but only countably many proofs.

The fact that there are more numbers than there are proofs doesn't mean anything, because a single proof can be about any number of numbers.  One proof can be a proof about a single number, a few numbers, countably infinitely many numbers, or an uncountable amount of numbers.
Example:
Claim: If $x+2 > 2$ then $x>0$.
Proof: Given $x+2>2$, subtract $2$ from both sides.  Then we get $x>0$. $\Box$
This is a (very simple) proof about an uncountable amount of numbers.  There are uncountably many numbers that are larger than $2$ (and larger than $0$).
A less trivial example of a proof that characterizes uncountably many numbers is the proof that there are uncountably many real numbers.
A: Take any unprovable theorem $T$ from your theory and define
$$\xi:=\begin{cases}T\text{ is true}\to +1,\\T\text{ is false}\to -1.\end{cases}$$
Can you prove $\xi>0$ ?
