The question goes as follows:
Prove that the following properties are all finitely productive
(1) $T_0$ and $T_1$
(2) Separable
(3) First Countable
(4) Second Countable
(5) Finite (i.e., the underlying set of the space is finite)
(6) Countable
(7) The property of being a discrete space
(8) The property of being an indiscrete space
Here's my proof;
Let $(X, \mathcal{T})$ and $(Y, \mathcal{U})$ be two topological spaces. We will prove finite productivity for two spaces. It generalizes well to any number of spaces.
(1) $T_0$ and $T_1$: Assume both $(X, \mathcal{T})$ and $(Y, \mathcal{U})$ are $T_0$ spaces, i.e., if $x_1, x_2$ and $y_1, y_2$ are points in $X$ and $Y$ respectively, then there exists open sets $U_1 \subseteq X$ and $U_2 \subseteq Y$ such that $U_1$ contains $x_1$ and not $x_2$ and $U_2$ contains $y_1$ and not $y_2$. Now consider the product space $X \times Y$. It is clear that $U_1 \times U_2 \subseteq X \times Y$ and that $x_1 \times y_1 \in U_1 \times U_2$ (since, $x_1 \in U_1$ and $y_1 \in U_2$). But $x_2 \times y_2 \notin U_1 \times U_2$. Thus $X \times Y$ is also $T_0$. Similarly, one can prove for $T_1$.
(2) Separable: Suppose $(X, \mathcal{T})$ and $(Y, \mathcal{U})$ are separable. Then, there exist countable dense subsets $D_X$ and $D_Y$ of $X$ and $Y$, respectively. We claim that $D_X \times D_Y$ is a countable dense subset of $X \times Y$. The countability is clear since both $D_X$ and $D_Y$ are countable $D_X \times D_Y$ must also be countable. Lets look at $\overline{D_X \times D_Y}$. We have $\overline{D_X \times D_Y} = \bar{D}_X \times \bar{D}_Y = X \times Y$. Thus $D_X \times D_Y$ is countable and dense in $X \times Y$. So, $X \times Y$ is separable.
(3) First Countable: Suppose $(X, \mathcal{T})$ and $(Y, \mathcal{U})$ are first countable, i.e., they have a local basis at each point. Let $x \in X$ and $y \in Y$ be two such points. Let $\mathcal{B}_x$ and $\mathcal{B}_y$ be the local basis at $x$ and $y$ respectively. Then we claim that $\mathcal{B}_x \times \mathcal{B}_y$ is a local basis at $x \times y$. To see this we verify the axioms of local basis at $x \times y$. The first axiom is automatically verified since, $x \times y \in \mathcal{B}_x \times \mathcal{B}_y$. To verify the second axiom, let $U$ and $V$ be two open sets containing $x$ and $y$ respectively. Then by hypothesis, there is a $B_1 \in \mathcal{B}_x$ and $B_2 \in \mathcal{B}_y$ such that $B_1 \subseteq U$ and $B_2 \subseteq V$. Then $B_1 \times B_2 \subseteq U \times V$ and $U \times V$ contains $x \times y$ from the first axiom of local basis. Thus, $X \times Y$ is first countable.
(4) Second Countable: Suppose $(X, \mathcal{T})$ and $(Y, \mathcal{U})$ are second countable, i.e., they have a countable basis $\mathcal{B}_X$ and $\mathcal{B}_Y$ respectively. From the definition of product topology, $\mathcal{B}_X \times \mathcal{B}_Y$ is a basis for $X \times Y$. Since, both $\mathcal{B}_X$ and $\mathcal{B}_Y$ are countable, the product $\mathcal{B}_X \times \mathcal{B}_Y$ is also countable. This proves that $X \times Y$ is second countable.
(5) Finite: Suppose both $X$ and $Y$ are finite. Then, $\vert X \times Y \vert = \vert X \vert\vert Y \vert$. Since both $\vert X \vert$ and $\vert Y \vert$ are finite, thus $\vert X \times Y \vert$ is also finite.
(6) Countable: Similarly we can prove countability as well.
(7) Discrete Space: If $(X, \mathcal{T})$ is discrete then, for $x \in X$, $\{x\}$ is both open and closed. Similarly for $(Y, \mathcal{U})$, we have $\{y\}$ to be closed for $y \in Y$. Since the product topology is generated by the basis $U \times V$ where $U \in \mathcal{T}$ and $V \in \mathcal{U}$, $\{x \times y\}$ must be open and closed in $X \times Y$. Thus $X\times Y$ is a discrete space.
(8) Indiscrete Space: Unable to prove this one.
I cannot prove the last bit of the question. Any help is appreciated. Thanks in Advance, Also feel free to criticize the rest of my proof.