I am not sure if this counter example works, but I am pretty sure. It is related to the counterexample mentioned in the question.

Consider in $\ell^1(\mathbb{Z})$ the usual ``base'' formed by the vectors $e_i$ given by the sequence having a $1$ in the position $i$.

We consider $G$ to be the subspace of $(\ell^1)^\ast = \ell^\infty$ generated by the dual of those vectors $e_i^\ast$ which clearly separates.

Now, consider $Y$ be the linear subspace generated by finite linear combinations of the vectors: $e_1$, $e_2 + 2e_3$, $e_3+ 2e_4+ 3e_5$, ....., $e_n + 2e_{n+1} + \ldots +(n-1) e_{2n-1}$, etc.

This subspace is dense since given any vector $v\in \ell^1$, we can arrange to construct a vector in $Y$ which coincides upto any finite number of coordinates with $v$.

However, if we consider for example the vector $x= \sum_i \frac 1 {n^2} e_i$, we can only aproach it with vectors in $Y$ of arbitrarily large norm.

I am not sure if this counter example works, but I am pretty sure. It is related to the counterexample mentioned in the question.

Consider in $\ell^1(\mathbb{Z})$ the usual ``base'' formed by the vectors $e_i$ given by the sequence having a $1$ in the position $i$.

We consider $G$ to be the subspace of $(\ell^1)^\ast = \ell^\infty$ generated by the dual of those vectors $e_i^\ast$ which clearly separates.

Now, consider $Y$ be the linear subspace generated by finite linear combinations of the vectors: $e_1$, $e_2 + 2e_3$, $e_3+ 2e_4+ 3e_5$, ....., $e_n + 2e_{n+1} + \ldots +(n-1) e_{2n-1}$, etc.

This subspace is dense since given any vector $v\in \ell^1$, we can arrange to construct a vector in $Y$ which coincides upto any finite number of coordinates with $v$.

However, if we consider for example the vector $x= \sum_i \frac 1 {i^2} e_i$, we can only aproach it with vectors in $Y$ of arbitrarily large norm.