Suppose $X$ follows a normal mixture with cdf:
$F(x)=0.5\Phi(\frac{x-\mu_1}{\sigma_1})+0.5\Phi(\frac{x-\mu_2}{\sigma_2})$,
where $\Phi(\cdot)$ denotes the normal cdf. Without further information, if I understand this correctly, $X$ does not necessarily follow a normal distribution. Is this correct?
When a finite mixture of Normal variables is Normal, then all the variables are identically distributed. Thus, if $\mu_1\ne\mu_2$ or $\sigma_1^2\ne\sigma_2^2,$ $F$ cannot be Normal.
A simple way to see why this assertion is true uses the characteristic function. Let the Normal components have variances $\sigma_1^2 \le \sigma_2^2 \le \cdots \le \sigma_n^2,$ corresponding means $\mu_1, \ldots, \mu_n,$ and nonzero proportions $\pi_1, \ldots, \pi_n.$ With no loss of generality, shift all means by a common amount to make the mixture mean zero. Suppose this mixture is Normal with the same zero mean and variance $\sigma.$ Its characteristic function must equal that of the mixture. Thus, for all real values $t,$
$$\exp(-\sigma^2 t^2/2) = \sum_{j=1}^n \pi_j \exp(i\mu_j t - \sigma_j^2 t^2/2).\tag{1}$$
Taking derivatives with respect to $t$ shows the left hand side is always real. The imaginary part of the derivative of the right side therefore must be constantly zero, especially in any neighborhood of $0$ where we may approximate the sine function by the linear term in its Taylor series:
$$0 = \frac{1}{2}\sum_{j=1}^n \pi_j \mu_j \sigma_j^2 \exp(- \sigma_j^2 t^2/2) \sin(\mu_j t) = \left(\frac{1}{2}\sum_{j=1}^n \pi_j \mu_j^2 \exp(-\sigma_j^2 t^2/2)\right) t + O(t^2).$$
This is impossible unless the coefficient of $t$ is identically zero, implying every $\mu_j=0.$ That is, all the component means must be equal if the mixture is Normal.
In light of this, $(1)$ asserts
$$\exp(-\sigma^2 t^2/2) = \sum_{j=1}^n \pi_j \exp(-\sigma_j^2 t^2/2) = \exp(-\sigma_1^2t^2/2 )\sum_{j=1}^n \pi_j \exp((\sigma_1^2-\sigma_j^2) t^2/2).$$
As $t^2$ grows large, the terms in the right hand sum either equal $\pi_j$ when $\sigma_j=\sigma_1$ or decrease rapidly to zero when $\sigma_j \gt \sigma_1.$ Taking logarithms, we find
$$\sigma^2t^2/2 = -\sigma_1^2 t^2/2 + \log\left(\sum_{j=1}^k \pi_j\right) + \epsilon,$$
where $\epsilon$ can be made arbitrarily small and $\sigma_1^2=\sigma_2^2=\cdots=\sigma_k^2 \ne \sigma_{k+1}^2.$ This is impossible unless the constant term on the right is zero, which means
$$1 = \sum_{j=1}^k \pi_j,$$
showing that $k=n.$ That is, all the $\sigma_j$ are equal, QED.
