The error in an alternating series that has been truncated to n terms is never any more than the absolute value of the next term.

So here if your series $\displaystyle \begin{align*} \frac{1 - \cos{(x)}}{x^2} = \frac{1}{2!} - \frac{x^2}{4!} + \frac{x^4}{6!} - \dots \end{align*}$ is truncated to two terms, then $\displaystyle \begin{align*} \frac{1 - \cos{(x)}}{x^2} \approx \frac{1}{2!} - \frac{x^2}{4!} \end{align*}$ with an error no greater than $\displaystyle \begin{align*} \frac{x^4}{6!} \end{align*}$.

So clearly $\displaystyle \begin{align*} \frac{1}{2!} - \frac{x^2}{4!} < \frac{1 - \cos{(x)}}{x^2} \end{align*}$, since the next term along is being added, thereby showing that your estimate is an UNDER estimate.

As for the other side of your inequality $\displaystyle \begin{align*} \frac{1}{2!} - \frac{x^2}{4!} < \frac{1 - \cos{(x)}}{x^2} < \frac{1}{2!} \end{align*}$, so where you are trying to show $\displaystyle \begin{align*} \frac{1 - \cos{(x)}}{x^2} < \frac{1}{2!} \end{align*}$ it is EXACTLY the same thing, just with your series truncated to one term. What would be the error then? Would the truncated series be an under estimate or an over estimate?