1. Let the roots of $x^2+ax+1=0$ be $\alpha$ and $\beta$.

2. For a quadratic equation $Ax^2+Bx+C=0$, the difference between the roots is $|\alpha-\beta|=\frac{\sqrt{B^2-4AC}}{|A|}.$

Here, $A=1$, $B=a$, $C=1$, so $|\alpha-\beta|=\sqrt{a^2-4}.$

3. Given that the difference between the roots is less than $\sqrt{5}$: $\sqrt{a^2-4}<\sqrt{5}.$ Squaring both sides, $a^2-4<5$ $a^2<9$ $-3<a<3.$

4. Also, since we are talking about the difference between the roots as a real quantity, the roots must be real. Hence, $a^2-4\ge 0 \quad \Rightarrow \quad |a|\ge 2.$ So strictly speaking, the values giving real roots are $a\in(-3,-2]\cup[2,3).$

5. However, among the given options, the intended inequality from $a^2<9$ corresponds to $(-3,3).$ So the matching option is:

$\boxed{\text{C}}$