Here is Prob. 10, Sec. 19, in the book Topology by James R. Munkres, 2nd edition:
Let $A$ be a set; let $\left\{ X_\alpha \right\}_{\alpha \in J}$ be an indexed family of spaces; and let $\left\{ f_\alpha \right\}_{\alpha \in J}$ be an indexed family of functions $f_\alpha \colon A \to X_\alpha$.
(a) Show there is a unique coarsest topology $\tau$ on $A$ relative to which each of the functions $f_\alpha$ is continuous.
(b) Let $$ \mathscr{S}_\beta = \left\{ \ f_\beta^{-1} \left( U_\beta \right) \ \vert \ U_\beta \mbox{ is open in } X_\beta \ \right\}, $$ and let $\mathscr{S} = \bigcup \mathscr{S}_\beta$. Show that $\mathscr{S}$ is a subbasis for $\tau$.
(c) Show that a map $g \colon Y \to A$ is continuous relative to $\tau$ if and only if each map $f_\alpha \circ g$ is continuous.
(d) Let $f \colon A \to \prod X_\alpha$ be defined by the equation $$ f(a) = \left( f_\alpha (a) \right)_{\alpha \in J}; $$ let $Z$ denote the subspace $f(A)$ of the product space $\prod X_\alpha$. Show that the image under $f$ of each element of $\tau$ is an open set of $Z$.
My Attempt:
Part (a):
The discrete topology on $A$ makes each function $f_\alpha$ continuous; so the collection of all those topologies on set $A$ relative to each one of which each one of $f_\alpha$ is continuous is non-empty, and the intersection $\tau$ of these topologies is the coarsest topology on $A$ relative to which each $f_\alpha$ is continuous.
Is my argument correct?
Part (b):
For each $\beta$, the set $X_\beta$ is open in $X_\beta$ and $f_\beta^{-1} \left( X_\beta \right) = A$ so that $A$ is in the collection $\mathscr{S}$. Thus $\mathscr{S}$ can be a subbasis for a topology on $A$; let $\tau^\prime$ be this topology. We show that $\tau^\prime = \tau$.
Since each function $f_\beta$ is continuous relative to $\tau$, therefore the set $f_\beta^{-1} \left( U_\beta \right)$ is in $\tau$ whenever $U_\beta$ is an open set in $X_\beta$. Thus $\mathscr{S}_\beta \subset \tau$ and hence $\mathscr{S} \subset \tau$, and as $\tau$ is a topology, so $\tau$ contains the intersections of all the finite subcollections of $\mathscr{S}$ and hence $\tau$ also contains all possible unions of these intersections; that is, $\tau$ contains all the sets in $\tau^\prime$. Therefore $\tau^\prime \subset \tau$.
Is my argument correct?
As $\mathscr{S}_\beta \subset \mathscr{S} \subset \tau^\prime$, so relative to the topology $\tau^\prime$ on $A$, each function $f_\beta$ is continuous; but $\tau$ is the coarsest of all such topologies on set $A$. So we can conclude that $\tau \subset \tau^\prime$. Hence $\tau^\prime = \tau$, as required.
Is my argument correct?
Part (c):
If the map $g \colon Y \to A$ is continuous relative to the topology $\tau$ on $A$, then as each map $f_\alpha \colon A \to X_\alpha$ is also continuous relative to the topology $\tau$, so each map $f_\alpha \circ g \colon Y \to X_\alpha$, being the composite of continuous functions, is also continuous.
Conversely, if each map $f_\alpha \circ g$ is continuous relative to topology $\tau$ on $A$, then for each open set $U_\alpha$ in $X_\alpha$, the inverse image $\left( f_\alpha \circ g \right)^{-1} \left( U_\alpha \right)$ is open in $Y$. But $$ \left( f_\alpha \circ g \right)^{-1} \left( U_\alpha \right) = g^{-1} \big( f_\alpha^{-1} \left( U_\alpha \right) \big). \tag{0} $$
Now let $V$ be any open set in $A$ relative to topology $\tau$. Then as $\mathscr{S}$ in Part (b) above is a subbasis for $\tau$, so our $V$ is a union of a collection of sets each of which is the intersections of some finite subcollection of $\mathscr{S}$; that is, $V$ is a union of sets, and each of those sets ( whose union is $V$ ) is the intersection of finitely many of the sets of the form $f_\alpha^{-1} \left( U_\alpha \right)$, where $\alpha \in J$ and $U_\alpha$ is open in $X_\alpha$.
As unions and intersections are preserved under $g^{-1}$, so $g^{-1}(V)$ is a union of sets each of which is an intersection of some finite subcollection of the following collection of sets: $$ \bigcup_{\alpha \in J} \big\{ \ g^{-1} \left( f_\alpha^{-1} \left( U_\alpha \right) \right) \ \colon \ U_\alpha \mbox{ is opne in } X_\alpha \ \big\} = \bigcup \big\{ \ \left( f_\alpha \circ g \right)^{-1} \left( U_\alpha \right) \ \colon \ U_\alpha \mbox{ is opne in } X_\alpha \ \big\} . \tag{1} $$ Here we have used (0) above.
As each of the sets in the collection (1) is open, so is any intersection of finitely many these sets and hence also any union of any of these intersections. Therefore $g^{-1}(V)$ is open in $Y$. Hence $g$ is continuous, as required.
Part (d):
Let $U$ be an open set in $A$ relative to $\tau$. We need to show that $f(U)$ is open in $f(Z)$, that is, there is an open set $V$ in $\prod X_\alpha$ such that $f(U) = f(Z) \cap V$.
Suppose $\mathbf{x} \in f(U)$. Then $\mathbf{x} = f(a)$ for some element $a \in U$ and so $\pi_\alpha ( \mathbf{x} ) = f_\alpha(a)$ for each $\alpha \in J$.
Now as $U$ is opne in $A$ and as $a \in U$, so there is a basis subset $U^\prime$ for the topology of $A$ such that $a \in U^\prime \subset U$, and also $U^\prime$ is the intersection of some finite subsollection of the subbasis $\mathscr{S}$ for $\tau$; that is, there are $\alpha_1, \ldots, \alpha_n \in J$ such that $$ U^\prime = \bigcap_{i=1}^n f_{\alpha_i}^{-1} \big( U_{\alpha_i} \big), \tag{1} $$ where $U_{\alpha_i}$ is open in $X_{\alpha_i}$ for each $i$. As for any $i = 1, \ldots, n$, and for any open sets $U_{\alpha_i}$ and $V_{\alpha_i}$ in $X_{\alpha_i}$, we note that $$ f_{\alpha_i}^{-1} \big( U_{ \alpha_i} \big) \cap f_{\alpha_i}^{-1} \big( V_{ \alpha_i} \big) = f_{\alpha_i}^{-1} \big( U_{ \alpha_i} \cap V_{ \alpha_i} \big), $$ which is again a set of the form which appears in the intersection on the right-hand-side of (1), so we can assume that the $\alpha_i$ in (1) are all distinct.
For any $\alpha \in J$ such that $\alpha \neq \alpha_i$ for any $i$, we can take $U_\alpha$ to be $X_\alpha$. Then (1) takes the form $$ U^\prime = \bigcap_{\alpha \in J} f_\alpha^{-1} \big( U_\alpha \big), \tag{2} $$ where $U_\alpha$ is open in $X_\alpha$ for each $\alpha \in J$ and $U_\alpha \neq X_\alpha$ for only finitely many $\alpha$.
Then $$ \mathbf{x} = f(a) \in f \left( U^\prime \right) = f \big( \bigcap_{\alpha \in J} f_\alpha^{-1} \big( U_\alpha \big) \big) \subset \bigcap_{\alpha \in J} f \big( f_\alpha^{-1} \big( U_\alpha \big) \big). \tag{3} $$
Also as $U^\prime \subset U$, so $$ f \left( U^\prime \right) \subset f(U). \tag{4} $$
Is what I have done so far correct? If so, then what next? How to proceed from here?
(d) is false, AFAIK. It certainly holds if the subbasic elements actually already form a base, which happens iff the family of functions $f_\alpha$ separates points from closed sets. (see here for the argument of that last fact. (i.e. for all $C \subseteq A$ closed and $x \notin C$ there is some $f_\alpha$ from our set of functions such that $f_\alpha(x) \notin \overline{f_\alpha[C]}$
So start looking for a counterexample... There is also some non-conclusive discussion here on that exact same subproblem.
For (c) just use that a map is continuous iff inverse images of subbasic elements are open. That's a bit shorter (and Munkres does prove this somewhere, so it's usable).
(a) and (b) are essentially OK, IMHO.