How to prove that $\xi_n + \eta_n \stackrel{d}\to \mathcal{N}(0, c)$ for some $c\,$? We have two sequences of random variables: $\xi_n \stackrel{d}\to \mathcal{N}(0, a)$, $\eta_n \stackrel{d}\to \mathcal{N}(0, b)$. Random variables are not necessarily independent.
How to prove that $\xi_n + \eta_n \stackrel{d}\to \mathcal{N}(0, c)$ for some $c$ ?
It is known that sum of two unbiased normal variables (probably dependent) is also normal (with some parameters), but how to transfer this fact to converging sequences?
 A: The result you refer to is not true in general; this is true, however, if consider a bivariate normally distributed pair $(\xi,\eta)$, but is not true in general. Consider for example $Z\sim\mathcal N(0,1)$ and $\mathbb P(U=1)=\frac12=\mathbb P(U=-1)$, where $U$ and $Z$ are independent. Then $UZ$ and and $Z$ are uncorrelated standard normal random variables, which are clearly dependent. Then $\mathbb P(UZ+Z=0)=\frac12$, so clearly there sum is not normal.
In general, if you can at least say that the vectors $(\xi_n,\eta_n)$ are i.i.d. in $\mathbb R^2$ with finite covariance matrix, then the multivariate CLT holds, hence by the continuous mapping theorem you can find a CLT for their sum. Note that this is also possible in the example I gave.
A: In general, this is not true. If $\left(X_n,Y_n\right) \xrightarrow{d} \left(X,Y\right)$, i.e., the convergence holds for $\left(X_n,Y_n\right)$ jointly, then you do have $X_n+Y_n\xrightarrow{d} X+Y$. However, if the convergence only holds separately for $X_n$ and $Y_n$, it does not need to hold for their sum.
One exception is the following case: if $X_n \xrightarrow{d} X$ and $Y_n\rightarrow c$ in probability for some constant $c$, then $X_n+Y_n\xrightarrow{d} X+c$.
