Metrizability of a topological vector space where every sequence can be made to converge to zero
It's true, at least for (DF)-spaces as proved by Kąkol and Saxon.
Let $X$ be a (DF)-space. Then the following are equivalent:
- $X$ admits a finer normed topology;
- $X$ has property $C_4$, that is, for any sequence $(x_n)_{n=1}^\infty$ in $X$, there is a sequence of positive reals $(t_n)_{n=1}^\infty$ s.t. $0\in \overline{\{x_n t_n\colon n\in \mathbb N\}}$.
It is not inconceivable that the requirement of having a finer normed topology can be replaced with having a weaker metric topology at the expense of dropping the hypothesis of being a (DF)-space.
For details, see Proposition 15.5 in
Jerzy Kąkol, Wiesław Kubiś, and Manuel López-Pellicer, Descriptive Topology in Selected Topics of Functional Analysis, Springer US, 2011.
For the counterexample below we need a little lemma about barrelled spaces (i.e., every barrel = closed, absolutely convex, and absorbing set is a $0$-neighbourhood):
If $(E,\mathcal T)$ is a barrelled locally convex space which has a finer metrizable vector space topology $\mathcal S$ then $(E,\mathcal T)$ is metrizable.
Indeed, let $(V_n)_{n\in\mathbb N}$ be a basis of the $0$-neighbourhoods in $(E,\mathcal S)$ und $U_n =\overline{\Gamma(V_n)}^{\mathcal T}$ the closed absolutely convex hulls of $V_n$. These are barrels in in $(E,\mathcal T)$ and hence $\mathcal T$-neighbourhoods of $0$. On the other hand, every $\mathcal T$-neighbourhood of $0$ contains a closed absolutely convex one which itself contains some $V_n$ (because $\mathcal S$ is finer) and hence some $U_n$. This means that $(U_n)_{n\in\mathbb N}$ is a countable basis of the $0$-neighbourhoods of $(E,\mathcal T)$ which is thus metrizable.
Now the counterexample: Let $I$ be an uncountable index set and $$E=\{(x_i)_{i\in I}\in \mathbb R^I: \{i\in I: x_i\neq 0\} \text{ is countable}\}$$ endowed with the relative topology of the product topology on $\mathbb R^I$. According to proposition 4.2.5(ii) in the book Barrelled Locally Convex Spaces of Bonet and Perez-Carreras, $E$ is barrelled. Since every $0$-neighbourhood only gives conditions on finitely many coordinates there is no countable basis of all $0$-neighbourhoods, i.e., $E$ is not metrizable and hence, by the lemma, it does not admit a finer metrizable vector space topology. On the other hand, for every sequence $x^n\in E$ the union $J$ of the supports of the $x^n$ is countable, so that $\{x^n:n\in\mathbb N\}$ is contained in the subspace $F=\mathbb R^J\times \{0\}^{I\setminus J}$ which is metrizable. Therefore, there is a sequence of $t_n>0$ such that $t_nx_n$ converges to $0$ in $F$ and hence also in $E$.
As you also asked for the particular case of strong duals of Frechet spaces $X$: This is not only covered by the answer of Tomasz Kania but it is rather trivial: If $X$ is not isomorphic to a Banach space there is a fundamental sequence of semi-norms $p_n$ which are pairwise non-equivalent. Then $X'=\bigcup_{n\in\mathbb N} X_n'$ where $X_n'$ is the space of functionals which are continuous with respect to $p_n$. Choosing $\phi_n\in X_n'\setminus X_{n-1}'$ there is no sequence of $t_n>0$ such that $t_n\phi_n\to 0$ because every bounded subset of $X'$ is contained in some $X_n'$.
EDIT. Another counterexample is the space of absolutely summable families $$\ell^1(I)=\{(x_i)_{i\in I}: \sum_{i\in I} |x_i|<\infty\}.$$ Every absolutely summable family has only countably many non-zero terms and hence a somehow natural topology (of course, besides the Banach space topology of the $\ell^1$-norm) is that of the inductive limit $\ell^1(I)=\lim\limits_\to \ell^1(J)$ for all countable subsets $J\subseteq I$. This finer topology is barrelled but not metrizable (hence it does not admit a finer metrizable vector space topology) but every sequence can be made bounded.