As others have stated, $\emptyset$ is a set with no elements and $\{\emptyset\}$ is a set with one element. Therefore these two sets can't be equal.
To address your specific question:
As for b. the empty set is empty therefore it can not include elements or be a subset (else from the empty set itself), is this claim concludes that $\{\emptyset\}\nsubseteq \emptyset$ or is it vacuous truth?
I think by "be a subset" you meant "have a subset," in which case, yes, that claim concludes that $\{\emptyset\} \nsubseteq \emptyset$. The reason I think you meant "have a subset" is because "be a subset" doesn't make sense in the context. The "subset" in question in this direction of the proof is $\{\emptyset\}$, which is the set containing the empty set. Although this set can potentially be a subset of another set, it can't be a subset of $\emptyset$ because $\emptyset$ has no elements. Therefore $\{\emptyset\}$ is not a subset of $\emptyset$.
To put it more concisely:
In general, $\{a\} \subseteq A$ if and only if $a \in A$.
So in our case, $\{\emptyset\} \subseteq \emptyset$ if and only if $\emptyset \in \emptyset$.
But of course $\emptyset \notin \emptyset$, because $\emptyset$ is empty.
Therefore $\{\emptyset\} \nsubseteq \emptyset$, and so $\{\emptyset\} \neq \emptyset$.