Does this constitute a valid proof that $\frac{x^2}{1+x^4} \leq \frac{1}{2}$?

Yes it's Ok, you could also just multiply both sides by $2(1+x^4)$ having to prove that $$ 2x^2\le x^4+1 $$ that is $$ 0\le(x^2-1)^2. $$

Start from $$x^4-2x^2+1=(x^2-1)^2\ge0$$ which is true for all $x\in\Bbb R$. Then rearrange: $$2x^2\le1+x^4$$ $$\frac{x^2}{1+x^4}\le\frac12$$ where the last step is done because 2 and $1+x^4$ are always positive for $x\in\Bbb R$.

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{x^{2} \over 1 + x^{4}} & = {1 \over 1/x^{2} + x^{2}} = {1 \over \pars{1/x - x}^{2} + 2} \color{#f00}{\leq \half} \end{align}

The equality is satisfied whenever $\ds{1/x - x = 0\quad\imp\quad x = \pm 1}$.