Locally countable implies countably tight? 
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*Locally countable: The space has a base of open sets, each with countable cardinality.

*Countably tight: for each A⊆X and each $p\in\overline{A}$ there is a countable subset $D\subseteq A$ such that $p\in \overline D$.

Does locally countable imply countably tight?

Yes, if $p \in \overline{A}$ and $U$ is open containing $p$, then $p \in \overline{U \cap A}$. So if $U$ is chosen to be countable, $U \cap A$ is as required for showing countable tightness.
