the problem posed is from Munkrees. Do you guys espouse my proof?? I feel as if I am missing something. Thank you!
Suppose that for each $i\in\mathbb{N}, X_i,$ and $Y_i$ are homeomorphic. Prove that $\displaystyle{\prod_{i\in\mathbb{N}} X_i}$ and $\displaystyle{\prod_{i\in\mathbb{N}} Y_i}$ are homeomorphic.
$\textbf{Proof:}$ Let $f_i: X_i \to Y_i$ be homeomorphism for all $i\in\mathbb{N}$. Then $f: \displaystyle{\prod_{i\in\mathbb{N}} X_i} \to \displaystyle{\prod_{i\in\mathbb{N}} Y_i}$ is the required homeomorphism. So, $f$ is continous because each $f_i$ is continous. Thereby, $f^{-1}: \displaystyle{\prod_{i\in\mathbb{N}} Y_i} \to \displaystyle{\prod_{i\in\mathbb{N}} X_i}$ which is equivalent to $(y_i)_{i\in\mathbb{N}}\to(f^{-1}_i(y_i))_{i\in\mathbb{N}}$. Thus, $f^{-1}$ is continous because each $f_i^{-1}$ is.
Your “proof” is no such thing. You just repeat the statement.
for a real proof, choose homeomorphisms $h_n: X_n \to Y_n$ for each $n$. From these construct $h: \prod_n X_n \to \prod_n Y_n$ in the obvious way, and use the universal property for maps into products to see that both $h$ and it’s inverse are continuous.
Or go high level and say that this follows from the fact that products in the category $\mathsf{Top}$ are limits and homeomorphisms are the category’s isomorphism. There is little topology in this fact.
added You have defined a sensible homeomorphism candidate. Depending on your text book and teacher demands you might have to add details to justify the claim that $f$ is continuous because all $f_n$ are (a theorem you can appeal to?) and maybe even on the bijectivity claim, depending on the amount of set theory covered.