weak* topology is not defined by any translation invariant metric when $X$ is infinite dimensional There is an exercise in Folland's real analysis, page 170

If $X$ is an infinite dimensional Banach space, then the weak* topology is not defined by any translation-invariant metric.

He gives a hint to this problem saying that 

Every weak*-Cauchy sequence in $X^*$ converges.

I don't know how to use his hint and the condition on "translation-invariance" to approach this problem. I know I should use the unboundedness of weak-open and weak*-open sets somewhere in the proof.
Furthermore, I wonder if the weak* topology can be defined by some metric that is not translation-invariant, which is a weaker result for this problem.
Thanks in advance.
 A: Since you are referring to the exercise in Folland's real analysis, I shall assume as facts the earlier parts of the problem, namely:

Every nonempty weak$^*$-open set in $X^*$ is unbounded.

I shall also assume the hint without proving it.
Now, assume on the contrary that $d$ is a metric that defines the weak$^*$ topology on $X^*$. Fix any $f_0 \in X^*$ and for $n=1,2,\dots$, pick $f_n $ in the weak$^*$-open ball $\{f:d(f,f_0)<1/n\}$ such that $||f_n||>n$ (this is possible by the unboundedness statement above).
Then $\{f_n\}$ is weak$^*$-Cauchy because given any $\epsilon > 0$, we can pick $N$ such that $2/N < \epsilon$, whence $d(f_n,f_m) \le d(f_n, f_0)+f(f_0, f_m) <2/N <\epsilon$ whenever $m,n \ge N$. By the hint, $f_n$ weak$^*$-converges to some $f$. That is, $f_n(x)\to f(x)$ for all $x\in X$. This contradicts the Uniform Boundedness Principle since we have $\sup_n |f_n(x)| < \infty$ for all $x\in X$ and yet $\sup_n || f_n|| = \infty$.
Remark: To answer the second question of OP, note that translation invariant is not used as part of the contrary statement.
