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Gerald Edgar
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Let's take the group $G = \mathbb Z^T$, where $T$ is an uncountable set.

Is $G$ weakly separable? Yes: If $U$ is any neighborhood of $0$ in $G$, then there exists a finite set $T_0 \subset T$ so that $$ U \supseteq V := \{\phi \in G \colon \phi(t)=0 \text{ for all } t \in T_0\} $$ Countably many translates of $V$ cover $G$: namely translates by all $\psi \in G$ supported by $T_0$.

If $G$ Lindelof? No. Steen & Seebach, Example 103.

Gerald Edgar
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