Let $X_n\sim\mathcal N_{\mu_n,\sigma_n^2}$ for some $(\mu_n,\sigma_n^2)\in\mathbb R\times(0,\infty)$ and $X$ be a real-valued random variable with $$X_n\stackrel{\text{in probability}}\to X\;.\tag 1$$
How can we prove that $X\sim\mathcal N_{\mu,\sigma}$ for some $(\mu,\sigma)\in\mathbb R\times(0,\infty)$?
That's what I've done: By $(1)$ and Slutsky's theorem we obtain $$X_n\stackrel{\text{in distribution}}\to X\tag 2$$ and thereby $$\operatorname E\left[f(X_n)\right]\stackrel{n\to\infty}\to\operatorname E\left[f(X)\right]\;\;\;\text{for all bounded }f\in C^0(\mathbb R)\tag 3$$ by the Portmanteau theorem. Especially, since $$\mathbb R\ni x\mapsto e^{{\rm i}x}=\cos x+{\rm i}\sin x$$ is continuous and bounded, $$\varphi_{X_n}(t):=\operatorname E\left[e^{{\rm i}tX_n}\right]\stackrel{n\to\infty}\to\operatorname E\left[e^{{\rm i}tX}\right]=:\varphi_X(t)\;\;\;\text{for all }t\in\mathbb R\tag 4\;.$$ $\varphi_{X_n}$ and $\varphi_X$ are the characteristic functions of $X_n$ and $X$, respectively. Now, we should be close since a finite measure on $\mathbb R^d$ is uniquely determined by its characteristic function.
However, I'm not able to conclude. Please note that I'm not primarily interested in $(\mu,\sigma)$ and the relationship to $(\mu_n,\sigma_n)$. I just want to show that $X$ is normally distributed. However, if it's easy to obtain, I wouldn't be angry if someone could prove the relationship too.
