I was confused about something really basic about Hausdorff distances and was wondering if someone could clarify.
Let $A \subset X $ be a non-empty set, and $a \in A$. Suppose further that $x \in X$.
Since the Hausdorff distance is a metric, it satisfies the triangle inequality. Can I say that
$$ d(x,a) \le d(x, A) +d(a, A) = d(x,A) \, ? $$ I know this isn't right. Where have I gone wrong? Can I not apply the triangle inequality like this, and if so, why can't I?
Assuming $a\in X, B \subset X$, I think you're getting confused between $$d(a,B) = \inf_{b \in B} d(a,b)$$ which isn't the Hausdorff metric, and $$d_H(\{a\}, B)$$ which is the only way I can see to interpret $d(a,B) $ as a Hausdorff metric.
You should re-check the definition, but unless $A = \{a\}$, $d_H(\{a\}, A)$ is not automatically zero. It's because, speaking roughly, there are points in $A$ that are some distance away from $a$.
Also, you're okay with singleton sets, so writing $d_H(\{x\}, \{y\}) $as $d(x,y)$ for your first term is valid, although likely to lead you astray. But once you dropped the "$d_H$" for "$d$" in the second half of the triangle inequality, you were doomed. :-)