 Solution

Weak convergence $X_k\Rightarrow X$ means that $\mathbb{E}f(X_k)\to \mathbb{E}f(X)$ for every bounded continuous real function $f$ on ${\mathbb{R}}^d$; it is a statement about their distributions, not their coupling.

The central limit theorem says that for independent identically distributed real variables of finite mean $\mu$ and variance ${\sigma }^2$, $\sqrt{n}({\bar{X}}_n-\mu )\Rightarrow N(0,{\sigma }^2)$. For $\sigma >0$ this is equivalent to convergence of $(\sum X_j-n\mu )/(\sigma \sqrt{n})$ to the standard normal distribution. For $\sigma =0$, every variable equals $\mu$ almost surely and the limit is the point mass at zero.

For the proof when $\sigma >0$, put $Y=(X-\mu )/\sigma$. Its characteristic function has expansion ${\phi }_Y(t)=1-t^2/2+o(t^2)$. To justify this using only a second moment, subtract the first two Taylor terms of $e^{itY}$ and use dominated convergence theorem, with the second-order remainder bounded by a constant times $t^2{Y}^2$. Independence gives the characteristic function of the normalized sum as $[{\phi }_Y(t/\sqrt{n}){]}^n\to e^{-t^2/2}$. The Lévy continuity theorem says that pointwise convergence of characteristic functions to a characteristic function continuous at zero implies weak convergence. Since $e^{-t^2/2}$ is the characteristic function of $N(0,1)$, the result follows.