What's your favorite proof accessible to a general audience? What math statement with proof do you  find most beautiful and   elegant, where such is  accessible to a general audience, meaning you could state, prove, and explain it to a general audience in roughly $5 \pm\epsilon$  minutes. 
Let's define 'general audience' as approximately  an average adult with education and experience comparable to someone holding a bachelor's degree in any non science major (e.g. history) from an average  North American university.
I really like the proof of $$\sum_{i=1}^n i = \dfrac{n(n+1)}{2}$$ in which $1 + 2 + \cdots + (n-1) + n$ is written forwards then backwards and summed. It is claimed that Gauss had come up with this when he was just a child, although contested.

The proof

Let $$s = 1 + 2 + \cdots + (n-1) + n.$$
Clearly,
$$ s = n + (n-1) + \cdots + 2 + 1.$$

Sum to get $$2s = \underbrace{(n+1) + (n+1) + \cdots  + (n+1)  + (n+1)}_{n \text{ times}}.$$

Hence, $$2s = n(n+1),$$ and $$s = \dfrac{n(n+1)}{2}.$$
One of the first proofs I was shown was in a discrete structures class.  I don't think this would survive the rigors of formal proof, but I really enjoyed it.

Definitions:

$N$ is the set of numbers starting with 0, 1, 2, $\dots $

A number in our set $N$ is odd if and only if there exists another number in our set $N$ that when multiplied by two and added to one yields the number in question.  Or more formally

$\forall a \in N$, a is odd if $\exists b \in N \ni a = 2b + 1$ 

To Prove:

If $n$ is odd then $n^2$ is odd 

Proof: Direct

Let us assume that we take an "odd" number, $n$.  Then "n is odd" is true.  Let us prove that $n^2$ is true.

By our assumptions $n$ is odd

This is the critical part of the proof, that might need some convincing

Since $n$ is odd we can use the definition of an odd number above and state that, $ \exists x \in N \ni n=2x+1$. 

Now we can substitute our number $n$ with $2x+1$ and state $n=2x+1$

Further to prove $n^2$ is odd, we need only show that $(2x+1)^2$ is odd

This also might need some examples to convince

By distributive property, $(2x+1)^2=4x^2+4x+1$

Now we can refactor and rewrite $4x^2+4x+1=2(2x^2+2x)+1$

We know from rules of arithimetic that $2x^2+2x$ is a number which we can call $z$ and $z=2x^2+2x$

By substitution $2(2x^2+2x)+1=$2z+1$

$\therefore$ by definition of odd number $2(2x^2+2x)+1$ is odd