In the text "Elementary Real Analysis The Theory of Calculus", I'm having trouble verifying my attempted proof that the function in $(1)$ is continuous via a $\delta$-$\epsilon$ approach.
Could anyone help to verify my proof, and/or point out errors I made, with suggestions to correct them?
$(1)\;\;$ Statement I need to prove: $\,f(x)$ is continuous for all $x$ in $\mathbb{R}$, where
$$f(x) = \sin^2(x) + \cos^6(x)$$
$\text{Lemma}:$ In order to unpack the notion that $f(x)$ is continuous for $x$ within $\mathbb{R}$, one needs the following theorem in $(2)$.
$(2)\;\;$
Let $f$ be a real-valued function whose domain is a subset $\mathbb{R}$. Then $f$ is continuous at $x_{o} \in \text{dom}(f)$ if and only if for each $\epsilon > 0$ there exists some $\delta > 0$ such that $|x-x_{o}| < \delta$ implies $|f(x)-f(x_{o})| < \epsilon$ for every $x \in \text{dom}(f)$.
$(3)\;\;$ Utilizing the recent results developed in $(2)$, one can make the following observations:
$$|f(x)-f(x_{o})| = |1-\cos^2(x) + \cos^2(x) - (1 - cos^2(x) + cos^{6}(x_{o})|$$ $$|1-cos^{2}(x) - 1-cos^2{(x_{o}})|$$ $$|1| \cdot |x - \cos^{2}({x_{o}})|$$
So it's clear that: $$|x - \cos^{2}({x_{o}})| < \frac{\epsilon}{|1|} \text{implies} \, |1| \cdot |x - \cos^{2}({x_{o}})| < |1| \cdot \frac{\epsilon}{|1|} = \epsilon$$