Prove that $\emptyset$ is unique initial object in Set(proof verification).

Question

First, we prove that the empty set is initial object in Set.

Suppose $f : \emptyset \rightarrow A$ is a function. Then, $f \subset \emptyset \times A = \emptyset$. Also, $\emptyset \subset f$ so $f = \emptyset \implies f = \emptyset$.

This shows:

$$Hom_{Set}(\emptyset,A) = \{\emptyset\}\ \forall A \in Obj(Set).$$

Thus, $\emptyset$ is initial object in Set. Suppose that B is initial. Then, $B \cong \emptyset \implies |B| = |\emptyset| = 0$. Since $\emptyset \subset B$ and $|B| = |\emptyset| < \infty$, therefore $B = \emptyset$.

Thus $\emptyset$ is unique initial object in Set.

Answer 1

The things you say are true. Perhaps, instead of using cardinality formalism for uniqueness, you could resort to a more elementary argument.

Let $L$ be an initial object. Then, there is (exactly) one morphism $f:L\to\emptyset$. Hence, $f\subseteq L\times\emptyset =\emptyset$. Therefore, $f=\emptyset$. But, by definition, $\emptyset$ is a function if and only if the domain is $\emptyset$. So $L=\emptyset$.