First of all, your proposed proof assumes what is to be proved. That kind of assumption can sometimes let you find a proof, but it can never be in a proof.

There is a fairly simple proof of this proposition if you know calculus.

$f(x) = \dfrac{x^2 - 2x + 4}{x^2 + 2x + 4} = 1 - \dfrac{4x}{x^2 + 2x + 4}.$

$\therefore f'(x) = -\ \dfrac{4(x^2 + 2x + 4) - (4x)(2x + 2)}{(x^2 + 2x + 4)^2} = \dfrac{-(4x^2 + 8x + 16) + (8x^2 + 8x)}{(x^2 + 2x + 4)^2} = \dfrac{4x^2 - 16}{(x^2 + 2x + 4)^2} \implies f'(x) = 0\ if\ x = \pm 2.$

$Also\ \displaystyle \lim_{x \rightarrow \infty}f(x) = 1 = \lim_{x \rightarrow - \infty}f(x).$

$f(2) = \dfrac{2^2 - 2(2) + 4}{2^2 + 2(2) + 4} \dfrac{4}{12} = \dfrac{1}{3}.$

$f(-2) = \dfrac{(-2)^2 - 2(-2) + 4}{(-2)^2 + 2(-2) + 4} \dfrac{12}{4} = 3.$

$\therefore \dfrac{1}{3} \le \dfrac{x^2 - 2x + 4}{x^2 + 2x + 4} \le 3.$

How you prove this without calculus is not readily apparent to me.