How would you prove this statement? Let $K$ be a subset of $\Bbb R$ (The real numbers)
Statement:
John likes $K$ if and only if $\exists a\in\Bbb R$ such that $\forall x\in K, a \le x$
Question: Does John like all subsets of the real numbers?
My answer: Yes he does, because I cannot think of a set that will prove this do be false. There always exits a real number which is going to be smaller than any real number in the set $K$.
How would I prove my answer?
 A: Consider the case where $K=\mathbb{Z}$. Can you find an $a$ such that $a$ is less than or equal to all of the integers?
A: $\bf HINT:$
Let me rephrase your statement a bit:
$P(K):$
$\exists a \in \Bbb R\forall x\in  {K} :a\leq x$.
Now, if $K= \Bbb R$, for the proposition to be true, we have to exhibit a real number that is smaller than any other real number, can you find such an element?
A: You can view these statements as a game. You go first; you choose a number $a\in\Bbb R$. Then, I choose a number $x\in K$. You win if $a\le x$; I win otherwise.
If you have a winning strategy for $K$, then John likes $K$.
Here's an example: $K=[0,1]$, the set of numbers between $0$ and $1$. On your turn, you choose $a=-1$. On my turn, no matter what $x$ I pick, we have $a\le x$ and you win! (Remember that $x\in K$.) So John likes $[0,1]$.
On the other hand, let $K=\Bbb R$. On your turn, you can pick anything — let's say you pick $a=-1000$. Then I can simply pick $x=-1001$, and you lose. In fact, there is no winning strategy for you! John doesn't like $\Bbb R$.
Other things that John doesn't like: $\Bbb Z$, $(-\infty,0]$, $\Bbb Q$, etc. Basically, John likes it iff it has a lower bound.
