I have found proof of the "delta method", (From Mathematical Statistics by Shao Jun P61) but I cannot understand some steps in this proof.
Theorem : Let $X_1, X_2,...$ and $Y$ be random k-vectors satisfying $$a_n(X_n-c)\to_dY$$ where $c\in\mathcal{R^k}$ and $\{a_n\}$ is a sequence of positive numbers with $\lim_{n\to\infty}a_n=\infty$. Let $g$ be a function from $\mathcal{R}^k\to\mathcal{R}$.
If $g$ is differentiable at $c$,then$$a_n[g(X_n)-g(c)]\to_d[\bigtriangledown g(c)]^TY$$ where $\bigtriangledown g(c)$ denotes the $k$-vector of partial derivatives of $g$ at $x$.
Proof:
Let$$Z_n=a_n[g(X_n)-g(c)]-a_n[\bigtriangledown g(c)]^T(X_n-c)$$
The differentiability of $g$ at $c$ implies that for any $\epsilon>0,\exists$ a $\delta_{\epsilon}>0$ s.t.$$|[g(X_n)-g(c)]-[\bigtriangledown g(c)]^T(X_n-c)|\le\epsilon\|x-c\|$$ whenever $\|x-c\|<\delta_{\epsilon}$. Let $\eta>0$ be fixed,$$P(|Z_n|\ge\eta)\le P(\|X_n-c\|\ge\delta_{\epsilon})+P(a_n\|X_n-c\|\ge\eta/\epsilon)$$ Since $a_n\to\infty$, $a_n(X_n-c)\to_dY$, Slutsky theorem implies $X_n\to_pc$(Why?)
Since $\|.\|$ is continuous, $a_n(X_n-c)\to_dY$ implies $$a_n\|X_n-c\|\to_d\|Y\|$$. WLOG, assume $\eta/\epsilon$ is a continuous point of $F_{\|Y\|}$. Then,$$\limsup_nP(|Z_n|\ge\eta)\le \lim_{n\to\infty}P(\|X_n-c\|\ge\delta_{\epsilon})+\lim_{n\to\infty}P(a_n\|X_n-c\|\ge\eta/\epsilon)=P(\|Y\|\ge\eta/\epsilon)$$
The proof is complete since $\epsilon$ can be arbitrary.(Why?)
(1) If $a_n W_n\xrightarrow{d} Y$, $$ W_n=\frac{a_n}{a_n}W_n\xrightarrow{d} 0\times Y=0, $$ which implies that $W_n\xrightarrow{p}0$.
(2) We can chose $\epsilon$ small enough (depending on $\eta$) such that $$ \limsup_{n\to\infty} \mathsf{P}(|Z_n|>\eta)<\eta, $$ which implies that $Z_n\xrightarrow{p}0$.
You can show (1) directly, i.e., for any $\eta>0$, $$ \limsup_{n\to\infty}\mathsf{P}(|W_n|>\eta)=\limsup_{n\to\infty}\mathsf{P}(a_n|W_n|>a_n\eta)\le \lim_{n\to\infty}\mathsf{P}(a_n|W_n|>y)<\eta. $$ where $y$ and $-y$ are continuity points of $F_Y$ s.t. $F_Y(-y)+(1-F_Y(y))< \eta$.