welcome to MSE.
The fact that strong convergence implies weak*-convergence in the dual space is so basic, that most books don't bother mentioning it. What might be a good thing to do is to reference "Functional Analysis" 2nd edition by Walter Rudin which is a good book on functional analysis containing a good introduction to the weak and weak* topology.
As for the proof, this consists of two parts, showing that strong convergence implies weak convergence and noting that weak convergence implies weak*-convergence.
Let $X$ be a normed vector space and let $X^{*}$ denote its dual space. By definition $X^{*}$ is the space of all continuous linear functional, so for any $\varphi\in X^{*}$ the operator norm
$$\|\varphi\|=\sup\{|\varphi(x)|:x\in X,\|x\|\leq1\}$$
is finite.
Now suppose $(x_{\alpha})$ is a net in $X$ (or $(x_{n})$ is a sequence if you don't want to use nets) converging strongly to $x\in X$. Then for all $\varphi\in X^{*}$ we find
$$\lim_{\alpha}|\varphi(x_{\alpha})-\varphi(x)|=\lim_{\alpha}|\varphi(x_{\alpha}-x)|\leq\lim_{\alpha}\|\varphi\|\|x_{\alpha}-x\|=0.$$
As this holds for all $\varphi\in X^{*}$ by definition $(x_{\alpha})$ weakly converges to $x$.
For the next part it is important to note that the weak* topology is not defined on $X$ but $X^{*}$. So we are considering strong convergence on $X^{*}$ with respect to the operator norm, and a net $(\varphi_{\alpha})$ in $X^{*}$ weakly converges to $\varphi$ if for all $\psi\in (X^{*})^{*}$ we have $\lim_{\alpha}\psi(\varphi_{\alpha})=\psi(\varphi)$. Recall that $(\varphi_{\alpha})$ converges to $\varphi$ in the weak* topology if for all $x\in X$ we have $\lim_{\alpha}\varphi_{\alpha}(x)=\varphi(x)$. Note however that for $x\in X$ the evaluation map $\psi_{x}:X^{*}\rightarrow\mathbb{R},\varphi\mapsto\varphi(x)$ is a continuous linear functional on $X^{*}$ hence $\psi_{x}\in(X^{*})^{*}$. Therefore if $(\varphi_{\alpha})$ converges to $\varphi$ in the weak topology it also converges in the weak* topology. (For a mathematical audience I would just write that the weak convergence implying weak* convergence is trivial.)
Combining these two results gives that strong convergence on $X^{*}$ implies weak* convergence.
Let me know if any of these steps is still unclear.
