Solution: Let $x \in \mathbb{R} $. Then, \begin{align*} (1 - x)^2 \geq 0 & \implies 1 + x^2 - 2x \geq 0\\ & \implies 1 + x^2 \geq 2x \\ & \implies 1 \geq \frac{2x}{1 + x^2} \\ & \implies \frac{x}{1 + x^2} \leq \frac{1}{2}. \end{align*} Thus, $\frac{1}{2}$ is an upper bound of the set $S$.

Note that for $x = 1$, \[ \frac{x}{1 + x^2} = \frac{1}{1 + 1} = \frac{1}{2} \in S. \] Thus, the upper bound $\frac{1}{2} \in S$ and hence it is the least upper bound of $S$. Therefore, \[ \textcolor{blue}{ \boxed{\mathrm{lub}(S) = \frac{1}{2} } } \]