The problem is: How to show that $\{N(\theta,1):\theta \in \Omega\}$ is not a complete family of distributions when $\Omega$ is finite?

For notational convenience, let $\varphi_{\theta}(x)$ denote the density of a $N(\theta, 1)$ random variable.

One way of viewing this problem is that the condition $$ \begin{align*} \int_{-\infty}^\infty g(x)\varphi_\theta(x)dx = 0, \quad \theta \in \{-1, 0, 1\} \tag{1} \end{align*} $$ sets up a system of three equations. If we pick up $g(x) = ax^3 + bx^2 + cx + d$ from the family of cubic polynomials, then $(1)$ becomes a homogeneous system of three linear equations with four unknowns $a, b, c, d$. By linear algebra theory, the dimensionality of the solution space to such a linear system is at least $1$ (because the rank of the associated coefficient matrix is at most $3$), implying that there are countless non-zero $g(x)$ satisfying $(1)$.

Indeed, substituting $g(x) = ax^3 + bx^2 + cx + d$ and central moments of $\varphi_\theta(x)$ into $(1)$ yields $$ \begin{align*} b + d &= 0, \\ 4a + 2b + c + d &= 0, \tag{2} \\ -4a + 2b - c + d &= 0. \end{align*} $$ One (of infinitely many) non-zero solution to $(2)$ is $a = -1, b = 0, c = 4, d = 0$, resulting $g(x) = -x^3 + 4x \neq 0$ (or more formally, $P_\theta(g(X) = 0) = P_\theta(X \in \{0, -2, 2\}) = 0 \neq 1$). This shows that the family $\{N(\theta, 1): \theta \in \Omega\}$ is not complete.

While the above construction works well for any parameter space with finite cardinality, it does not generalize to the case where the parameter space contains infinite members (e.g., the original linked exercise whose $\Omega = \{1, 2, 3, \ldots\}$). To deal with the latter case, note that, by the translation property of $\varphi_\theta$, the condition $E_\theta[g(X)] = 0$ for all $\theta \in \Omega$ is equivalent to
$$ \begin{align*} E[g(X + \theta)] = 0 \quad \text{ for all } \theta \in \Omega, \tag{3} \end{align*} $$ where “$E$” is with respect to the standard normal density $\varphi(x)$. This observation implies that, when $\Omega = \{1, 2, \ldots\}$, $(3)$ automatically holds for $g$ such that $g(x + \theta) = g(x)$ for all $\theta = 1, 2, \ldots$ and $E[g(X)] = 0$. Clearly, one candidate of such $g$ is any periodic odd function with periodicity $1$. For example, $g(x) = \sin(\pi x)$ proposed in @jld’s answer. As another (non-trigonometric) example, $g(x)$ can be taken to be $$ \begin{align*} g(x) = \frac{1}{2} - \left|x - \frac{1}{2}\right|, \quad 0 \leq x \leq 1. \end{align*} $$ And then copy this to intervals $[n, n + 1]$, $n = 1, 2, \ldots$ to complete $g$’s definition on $[0, +\infty)$. Finally, define $g(x) = -g(-x)$ when $x < 0$.