Banach-Alaoglu states that: If $X$ is a topological vector space then the polar of any neighborhood of the origin is $\sigma(X^*,X)$ compact. Especially, if $X$ is a norm linear space then the closed unit ball $B^*$ in $X^*$ is $w^*$ compact. From this, since for $a\in X$ and $\lambda\neq 0$, the two mappings $x\mapsto x+a$ and $ x\mapsto \lambda x$ are homogeneous, we can conclude that each ball $\mathbb B^*(x^*;r)$ is also $w^*$ compact. Now, suppose that $C$ is a closed, bounded, and convex subset in $X^*$ then $C$ is contained in some closed ball. This implies, if $C$ is also $w^*$ closed then $C$ must be $w^*$ compact. But closed and convex just implies $C$ is weakly closed, i.e., $\sigma(X^*, X^{**})$ closed.
I would like to show that there exists a closed, bounded and convex subset $C$ in $X^*$ but $C$ even not $w^*$ closed.
This example show that in some non-reflexive $X$, topology $\sigma(X^*,X)$ is stricly weaker than topology $\sigma(X^*,X^{**})$ and there are some closed and convex subset $C$ in $X^*$ which can not be the polar of any neighborhood of the origin in $X$.
I tried to find that set in $X^*=\ell^1$ of $X=c_0$ given by $$C=\{f=(f_i)_i:\; \; \sum_{n=1}^\infty f_i =1, f_i \geq 0 \; \mbox{ for all } \; i\}.$$
(1) It is obvious that $C$ is convex and bounded. (2) Now, take $f^{(n)}=(0, \ldots, 0, 1_n, 0,\ldots)\in C$. For any $x=(x_i)_i\in X=c_0$: $$f^{(n)}(x)=\sum_{i=1}^\infty x_if_{i}^{(n)}=x_n\to 0 \; \mbox{ as }\; n\to \infty.$$ Thus $f^{(n)} \to^{w^*} 0$ but $0\notin C$ and hence $C$ is not $w^*$ closed.
It remains to show that $C$ is closed by norm in $X^*$. Thanks in advance.
Your set $C$ is closed. Consider bounded linear functionals $$ {\rm ev}_i:\ell_1\to\mathbb{C}:f\mapsto f_i\\ g:\ell_1\to\mathbb{C}:f\mapsto \sum_{i=1}^\infty f_i $$ then, the sets ${\rm ev_i}^{-1}([0,+\infty))$, $g^{-1}(\{1\})$ are closed. Since $$ C=g^{-1}(\{1\})\cap\left(\bigcap\limits_{i=1}^\infty {\rm ev}_i^{-1}([0,+\infty))\right) $$ then $C$ is closed as intersection of closed sets.
I can suggest another example. Consider bounded linear functional $$ f:\ell_1\to\mathbb{C}:x\mapsto\sum\limits_{n=1}^\infty\frac{n}{n+1}x(n) $$ Clearly, $\ker f$ is a closed subspace, hence $C=B_{\ell_1}\cap\ker f$ is convex closed and bounded. We claim it is not not weak-${}^*$ closed. Indeed the sequence $(x_n)_{n\in\mathbb{N}}\subset C$ where $x_n=\frac{n}{2n+1}e_1-\frac{n+1}{2n+1}e_n$ weak-${}^*$ converges to $\frac{1}{2}e_1\notin C$