Examples are not proofs.

Consider the following problem.

For any $x\in \mathbb{R}$, prove that

$$-|x|\le x\le |x|.$$

In case you forgot, the absolute value of a real number $x$ is defined as follows:

$$|x|=\{\begin{array}{cc}x& \text{if }x\ge 0,\\ -x& \text{if }x<0.\end{array}$$

A notable property of the absolute value is that it is always non-negative (positive or zero).

Proof (wrong!): Let $x=3$. Then $|x|=3$. Thus, we have 

$$-3\le 3\le 3.$$ 

So the proposition is true. Let $x=-3$. Then $|x|=|-3|=3$. Thus, we have

$$-3\le -3\le 3.$$

So the proposition is also true. In either case, the proposition is true. Proved! ■

What is exactly wrong with this "proof"? It is that it only proves the cases when $x=3$ or $x=-3$. But we need to prove that for any real number $x$, and there are infinitely many (in fact, uncountably many) real numbers! What about $x=2.3$, or $x={\pi }^{-\sqrt{2}}$, or ...? Unless we prove the proposition for the general case, there is a possibility that the proposition might fail to hold for some strange real numbers.

Proof (correct): Let $x$ be an arbitrary real number. If $x\ge 0$, then $|x|=x$. Thus,

$$-|x|\le 0\le x\le |x|.$$

(Recall that $a\le b$ means $a<b$ or $a=b$.)

If $x<0$, then $|x|=-x$. Thus,

$$-|x|\le x<0\le |x|.$$

In either case, the proposition is true. ■

In this proof, $x$ is an arbitrary real number, and what we show here applies to any real number. Therefore, the proof is general and correct.