From your specified model you have likelihood function:
$$L_\mathbf{x}(\mu_1,\mu_2) = (8 \pi)^{-n/2} \prod_{i=1}^n \Bigg[ \exp \Big( -\frac{1}{2} (x_i-\mu_1)^2 \Big) + \exp \Big( -\frac{1}{2} (x_i-\mu_2)^2 \Big) \Bigg].$$
To facilitate the analysis, let $\mathscr{B} \equiv \{ 1,2 \}^n$ be the set of all $n$-length binary vectors of values of the two means. For any vector $\mathbf{b} \in \mathscr{B}$ let $\mathcal{S}(\mathbf{b}) \equiv \{ i=1,...,n | b_i=1 \}$ be the set of index values that use the first mean. With the improper uniform prior $\pi (\mu_1, \mu_2) \propto 1$ you get the posterior:
$$\begin{equation} \begin{aligned}
\pi (\mu_1, \mu_2 | \mathbf{x})
&\propto L_\mathbf{x}(\mu_1,\mu_2) \\[6pt]
&\propto \prod_{i=1}^n \Bigg[ \exp \Big( -\frac{1}{2} (x_i-\mu_1)^2 \Big) + \exp \Big( -\frac{1}{2} (x_i-\mu_2)^2 \Big) \Bigg] \\[6pt]
&= \sum_{\mathbf{b} \in \mathscr{B}} \exp \Big( -\frac{1}{2} \sum_{i=1}^n (x_i-\mu_{b_i})^2 \Big) \\[6pt]
&= \sum_{\mathbf{b} \in \mathscr{B}} \exp \Big( -\frac{1}{2} \sum_{i \in \mathcal{S}(\mathbf{b})} (x_i-\mu_1)^2 \Big) \exp \Big( -\frac{1}{2} \sum_{i \notin \mathcal{S}(\mathbf{b})} (x_i-\mu_2)^2 \Big). \\[6pt]
\end{aligned} \end{equation}$$
For the vectors $\mathbf{b} = (1,...,1)$ and $\mathbf{b} = (2,...,2)$ we get empty classes over one of the inner sums and so the outer summation includes one term that does not depend on $\mu_1$ and one term that does not depend on $\mu_2$. For all other terms (i.e., with some values from each mean) we can simplify as follows. Letting $\bar{x}_1^{(k)} (\mathbf{b}) \equiv \sum_{i \in \mathcal{S}(\mathbf{b})} x_i^k$ and $\bar{x}_2^{(k)} (\mathbf{b}) \equiv \sum_{i \notin \mathcal{S}(\mathbf{b})} x_i^k$ we have:
$$\begin{equation} \begin{aligned}
\exp \Big( -\frac{1}{2} \sum_{i \in \mathcal{S}(\mathbf{b})} (x_i-\mu_1)^2 \Big)
&= \exp \Big( -\frac{1}{2} \Big( \mu_1^2 + 2 \mu_1 \sum_{i \in \mathcal{S}(\mathbf{b})} x_i + \sum_{i \in \mathcal{S}(\mathbf{b})} x_i^2 \Big) \Big) \\[6pt]
&= \exp \Big( -\frac{1}{2} ( \mu_1^2 + 2 \mu_1 \bar{x}_1^{(1)} (\mathbf{b}) + \bar{x}_1^{(2)} (\mathbf{b}) ) \Big) \\[10pt]
&= \exp \Big( -\frac{1}{2} ( \bar{x}_1^{(2)} (\mathbf{b}) - \bar{x}_1^{(1)} (\mathbf{b})^2 ) \Big)
\exp \Big( -\frac{1}{2} (\mu_1 - \bar{x}_1^{(1)} (\mathbf{b}))^2 \Big) \\[10pt]
&\propto \exp \Big( -\frac{1}{2} ( \bar{x}_1^{(2)} (\mathbf{b}) - \bar{x}_1^{(1)} (\mathbf{b})^2 ) \Big) \cdot \text{N}(\mu_1 | \bar{x}_1^{(1)} (\mathbf{b}), 1), \\[10pt]
\exp \Big( -\frac{1}{2} \sum_{i \notin \mathcal{S}(\mathbf{b})} (x_i-\mu_2)^2 \Big) &\propto \exp \Big( -\frac{1}{2} ( \bar{x}_2^{(2)} (\mathbf{b}) - \bar{x}_2^{(1)} (\mathbf{b})^2 ) \Big) \cdot \text{N}(\mu_2 | \bar{x}_2^{(1)} (\mathbf{b}), 1). \\[10pt]
\end{aligned} \end{equation}$$
Let $\bar{\mathscr{B}}$ denote the vectors that have mixtures of values from both parts of the mixture (i.e., excluding $\mathbf{b} = (1,...,1), (2,...,2)$). Let $H(\mathbf{b}) \equiv \bar{x}_1^{(2)} (\mathbf{b}) - \bar{x}_1^{(1)} (\mathbf{b})^2 + \bar{x}_2^{(2)} (\mathbf{b}) - \bar{x}_2^{(1)} (\mathbf{b})^2$ and we then have:
$$\begin{equation} \begin{aligned}
\pi (\mu_1, \mu_2 | \mathbf{x})
&\propto \exp \Big( -\frac{1}{2} \sum_{i=1}^n (x_i-\mu_1) \Big) + \exp \Big( -\frac{1}{2} \sum_{i=1}^n (x_i-\mu_2) \Big) \\[6pt]
&+ \sum_{\mathbf{b} \in \bar{\mathscr{B}}} \exp \Big( -\frac{1}{2} H(\mathbf{b}) \Big) \cdot \text{N}(\mu_1 | \bar{x}_1^{(1)} (\mathbf{b}), 1) \cdot \text{N}(\mu_2 | \bar{x}_2^{(1)} (\mathbf{b}), 1). \\[6pt]
\end{aligned} \end{equation}$$
This kernel has an infinite integral, owing to integration over the terms that do not depend on both mean parameters. Hence, the posterior is improper. (Thanks to Xi'an for pointing out an error in the previous version of this answer.)