There is a another way to solve this.
We know that $\gamma$ is a valid autocovariance function iff, for every $n > 0$, the matrix $\Gamma_n$ defined as $\Gamma_n(i,j) = \gamma(|i-j|)$ is positive semidefinite.
In particular we must have that, for every $n > 0$, $\sum_{i,j} \Gamma_n(i,j) \geq 0$
Now let's assume $\gamma(h) = 0$ for all $h$ such that $|h|>3$ and $\gamma(0)=1$ (this includes your case).
Then for $n \geq 3$,
$$
\sum_{i,j} \Gamma_n(i,j) = n (1 + 2 \gamma(1) + 2 \gamma(2) + 2 \gamma(3)) + C
$$
where $C$ does not depend on $n$.
Since this has to be non-negative for every $n$, so must it be its limit for $n \to \infty$.
This implies that
$$
\gamma(1) + \gamma(2) + \gamma(3) \geq - \frac 12
$$
which is not satisfied in your case
(which is $\gamma(1) = 0, \gamma(2) = -0.5, \gamma(3) = -0.25$)