Here is Prob. 2, Sec. 29, in the book Topology by James R. Munkres, 2nd edition:
Let $\left\{ \ X_\alpha \ \right\}$ be an indexed family of nonempty spaces.
(a) Show that if $\prod X_\alpha$ is locally compact, then each $X_\alpha$ is locally compact and $X_\alpha$ is compact for all but finitely many values of $\alpha$.
(b) Prove the converse, assuming the Tychonoff theorem.
My Attempt:
Prob. 2 (a):
Let us put $$ X \colon= \prod X_\alpha. \tag{Definition 0}$$ Suppose that $X$ is locally compact.
Let us take a particular index $\alpha_0$. We show that $X_{\alpha_0}$ is locally compact. Let $p_{\alpha_0}$ be any point of $X_{\alpha_0}$. Let us choose some point $\mathbf{p} \in X$ such that $$ \pi_{\alpha_0} ( \mathbf{p} ) = p_{\alpha_0}, \tag{0} $$ where $\pi_{\alpha_0}$ is the projection mapping of the product $X = \prod X_\alpha$ onto the factor $X_{\alpha_0}$.
Now as $X$ [Refer to (Definition 0) above.] is locally compact at $\mathbf{p}$, so there exists a compact subspace $C$ of $X$ containing a neighborhood $U$ of $\mathbf{p}$ in $X$; that is, $$ \mathbf{p} \in U \subset C \tag{1} $$ holds, where $U$ is an open set in $X$ and $C$ is a compact subspace of $X$.
Since the projection map $\pi_{\alpha_0}$ is an open and continuous map of $X = \prod X_\alpha$ into (in fact onto) $X_{\alpha_0}$ and since $U$ is an open set of $X$ and $C$ is a compact subspace of $X$, therefore the image set $\pi_{\alpha_0}(U)$ is an open set in $X_{\alpha_0}$ and the image set $\pi_{\alpha_0}(C)$ is compact (in fact a compact subspace of $X_{\alpha_0}$).
Moreover (0) and (1) above together imply that $$p_{\alpha_0} = \pi_{\alpha_0} ( \mathbf{p} ) \in \pi_{\alpha_0}(U) \subset \pi_{\alpha_0}(C). $$ Moreover, $\pi_{\alpha_0}(U)$ is open in $X_{\alpha_0}$ and $\pi_{\alpha_0}(C)$ is compact. Therefore $X_{\alpha_0}$ is locally compact at any point $p_{\alpha_0} \in X_{\alpha_0}$, as required. [Refer to (Definition 0) above.]
Furthermore, as $U$ is an open set of $X$ containing point $\mathbf{p}$ [Refer to (1) above.], so there exists a canonical basis set $B$ for the product topology on $X$ such that $$ \mathbf{p} \in B \subset U. $$ But from (1) we have $U \subset C$. So we also obtain $$ B \subset C, \tag{2*}$$ and hence $$ \pi_\alpha (B) \subset \pi_\alpha(C) \tag{2} $$ for each index $\alpha$.
But by definition of the product topology, the set $B = \prod B_\alpha$, where each set $B_\alpha$ is an open set in the corresponding topological space $X_\alpha$ and $B_\alpha$ equals $X_\alpha$ except for some finitely many values of $\alpha$ only. Thus (2*) becomes $$ \prod B_\alpha \subset C. \tag{2**} $$ So from (2**) and (2) we can conclude that the image set $\pi_\alpha(C)$ of the set $C$ under the projection map $\pi_\alpha$ of $X$ onto $X_\alpha$ equals $X_\alpha$ itself except only for some finitely many values of $\alpha$.
But as each one of the projection maps $\pi_\alpha \colon X \to X_\alpha$ is continuous and as $C$ is a compact subspace of $X$, so each image set $\pi_\alpha(C)$ is also compact.
From the preceding two paragraphs we can conclude that each space $X_\alpha$ is compact except only for some finitely many values of $\alpha$, as required.
Is this proof correct and clear enough in each and every bit?
Prob. 2 (b):
Now we suppose that, except for at most a finite number of values of the index $\alpha$, the spaces $X_\alpha$ are all compact, and we also suppose that even those of the spaces $X_\alpha$ that are not compact are at least locally compact. Remember that every topological space that is compact is also locally compact but not every locally compact space is compact. Let $\alpha_1, \ldots, \alpha_n$ be the values of $\alpha$ for which the spaces $X_\alpha$ are only locally compact.
Let us put again $$ X \colon= \prod X_\alpha. \tag{Definition 0} $$
Let $\mathbf{p} \colon= \left( p_\alpha \right)$ be any point of $X$. We show that $X$ is locally compact at $\mathbf{p}$.
For each $j = 1, \ldots, n$, as $p_{\alpha_j} \in X_{\alpha_j}$ and as $X_{\alpha_j}$ is locally compact (at $p_{\alpha_j}$), so there exists a compact subspace $C_{\alpha_j}$ of $X_{\alpha_j}$ containing a neighborhood $U_{\alpha_j}$ of $p_{\alpha_j}$ in $X_{\alpha_j}$, that is we have $$ p_{\alpha_j} \in U_{\alpha_j} \subset C_{\alpha_j}, \tag{0} $$ where $U_{\alpha_j}$ is open in $X_{\alpha_j}$ and $C_{\alpha_j}$ is a compact subspace of $X_{\alpha_j}$.
Now let us put $$ U \colon= \prod U_\alpha, \tag{Definition 1} $$ where $$ U_\alpha \colon= X_\alpha \ \mbox{ if } \ \alpha \neq \alpha_1, \ldots, \alpha_n, \tag{Definition 1}$$ and let us put $$ C \colon= \prod C_\alpha, \tag{Definition 2} $$ where $$ C_\alpha \colon= X_\alpha \ \mbox{ if } \ \alpha \neq \alpha_1, \ldots, \alpha_n. \tag{Definition 2}$$ The $U_\alpha$ and the $C_\alpha$ for $\alpha = \alpha_1, \ldots, \alpha_n$ are as in (0) above in the preceding paragraph of course.
Then from (0) above we can conclude that $$ \mathbf{p} \in U \subset C. $$ Moreover, $U$ is an open set of $X$ [Refer to (Definition 0) above.] In fact $U$ is a set in the canonical basis for the product topology on $X$, and by the Tychonoff theorem the set $C$ is a compact subspace of $X$.
Thus $U$ is a neighborhood of $\mathbf{p}$ in $X$ and $C$ is a compact subspace of $X$ containing $U$. Therefore $X = \prod X_\alpha$ is locally compact at any point $\mathbf{p} \in X$, as required.
Is this proof good enough? If so, is the presentation clear enough too? Or, are there problems?