For the RHS inequality you can even use Mean Value Theorem (which, of course, is a special case of Taylor theorem - where the reminder term contains first derivative).
Let $f(x) = \sqrt{1+x}$. Then the theorem gives that there exists some $\theta \in (0,x)$ such that \frac{f(x)-f(0)}{x-0} = \frac{\sqrt{1+x} - 1}{x} = f'(\theta) = \frac{1}{2 \sqrt{1+\theta}}. Hence $\sqrt{1+x} = 1 + \frac{x}{2 \sqrt{1+\theta}} < 1 + \frac{x}{2}$ since $x > 0$. Observe that, of course, $\frac{x}{2 \sqrt{1+\theta}}$ is the reminder term in the Taylor theorem.
For the LHS by Taylor theorem you get that there exists $\theta \in (0,x)$ such that $ \sqrt{1+x} = 1 + \frac{1}{2} x - \frac{1}{8} x^2 + \frac{1}{16} (1 + \theta)^{-5/2} x^3. $ The reminder term is positive for $x > 0$, which proves the inequality.