Showing that $x$ is non-positive I've just started to study analysis by myself and I'm having a hard time proving things. I can understand the logic behind it and can describe it by words but I can't write it in "mathematics". This is the question:
$x, y ∈ R$ 
If for all $y>0$ we have $y≥x$ show that $0≥x$
What I understand is that basically it says if all numbers from 0 to positive infinity is bigger than x than x is either 0 or a negative number because even the small numbers like $1/{10^{100000000}}$ are bigger than x. But how can I prove this  in math? It seems pretty straightforward.
 A: 
I can understand the logic behind it and can describe it by words but I can't write it in "mathematics". This is the question:

I just want to disabuse you of a belief here, that one has to write an argument using symbolic notation (which is how I interpret writing in mathematics) before it is a valid mathematical proof. This is not true. Some (indeed, most) mathematical arguments are more conveniently stated and understood when there are abbreviations, but that does not mean all arguments must be like this. Indeed, some arguments are better stated literally, as they say. It would be overkill to use anything but words to describe very primitive arguments (by which I mean they are closer to the axioms than most, by which I mean the chain of deduction therefrom is short or only an inference away).
Now I should not be misunderstood. That one is stating a proof (as opposed to a heuristic or informal argument, say, which also has its place) literally does not mean one should be sloppy or cavalier; hence one still has to follow accepted standards of mathematical logic, for example, be precise in one's use of words, and be clear. (Indeed, it was to enhance these effects in some proofs that symbolic notation was invented, and its usefulness becomes most evident then -- it just becomes near impossible to present some arguments in usual language to a satisfactory level of rigour, clarity and precision, which are the life blood of mathematics).

What I understand is that basically it says if all numbers from 0 to positive infinity is bigger than x than x is either 0 or a negative number...

The key here is the word all. It is then immediately obvious that there can be no place for $x$ anywhere in the positive real axis since all positive real numbers $y$ are greater than $x.$ If someone were to press us further, then we can do no better than show that a contrary assumption leads to a contradiction; for suppose $x$ was positive, then that would mean $x$ was greater than some positive real number $y,$ which is contrary to our hypothesis.
Usually, though, the explanation by contradiction is no more logically prior than just noting that the constraints make it impossible -- it all depends on whether our starting axioms agree. Thus the argument by contradiction makes use of the fact that $\mathbf R$ is dense (in itself), for example, which leads to other questions, ad infinitum. Eventually, we have to stop somewhere in our explanation and take a set of "facts" as basic -- these we call axioms (and what's obvious and basic to one may not be to another).
In all, to prove something, it is not a requirement that you use any non-usual (non-literal, as in normal written language) symbol, as long as it satisfies the other conditions of logical validity, clarity and precision.
Welcome to analysis and good luck with your studies.
A: You know that $x$ is less than or equal every positive number and you want to show that $x$ is not positive. 
Well, if $x$ is positive then it must be less than or equal $x/2$ which is not possible, so $x$ is not positive.
Thus $x\le 0$  
A: Suppose
$x \le 0 \tag 1$
is false.  Then
$x > 0; \tag 2$
set
$y = \dfrac{1}{2} x; \tag 3$
then
$0 < y < x. \tag 4$
We have just shown the existence of $y \in \Bbb R$ with $y > 0$ and $y < x$, which contradicts the hypothesis
$\forall 0 < y \in \Bbb R, \; y \ge x; \tag 5$
therefore there is no such $x > 0$; thus (1) binds.
A: If $x>0$, take $y=\frac x2$. Then $y>0$, too. Besides, $x>y$. This is impossible, because $y>0$ and therefore $y\geqslant x$.
Since we reached a contradiction, the original assumption ($x>0$) is false. Therefore, $x\leqslant0$.
