Below is a question from Topology, James Munkres. Following that is my attempt at a solution, which I am not sure is correct and would appreciate if somebody could point out what (if anything) is wrong with it.
$\textbf{Question:}$ Show that the product of an uncountably many copies of the real line is not metrizable.
$\textbf{Attempted Solution:}$ Let $X$ denote the space $\prod_{\alpha\in J} \mathbb{R}_{\alpha}$. In order to show that $X$ is not metrizable, it suffices to show that $X$ is not first-countable.
Let $X$ be first-countable. Then given a point $x=(x_{j})_{j\in J}$ in $X$, there exists a countable collection $\{U_{n}\}_{n\in N}$ of open neighbourhoods of $x$ such that every open set containing $x$ contains atleast one of the $U_{j}$. Let $V_{j}$ denote the basis element about $x$ contained in $U_{j}$. Then $V_{j}$ is of the form $\pi^{-1}_{\alpha_{1_{j}}}(W_{\alpha_{1_{j}}})\cap \pi^{-1}_{\alpha_{2_{j}}}(W_{\alpha_{2_{j}}})\cap \dots\pi^{-1}_{\alpha_{n_{j}}}(W_{\alpha_{n_{j}}})$. Let $I$ be the subset of $J$ consisting of the indices $\alpha_{i_{j}}$ for all $j\in \mathbb{N}$ and $1<i<n$. Then $I$ is a countable union of finite sets, so is countable and therefore $J-I$ is nonempty.
For some $\gamma\in J-I$ and some open neighbourhood $V_{\gamma}\subsetneq X_{\gamma}$ containing $\pi_{\gamma}(x)$, let $\pi^{-1}_{\gamma}(V_{\gamma})$ be an open set containing $x$. Then $\pi^{-1}_{\gamma}(V_{\gamma})$ must contain an open set of the form $U_{k}$, which in turn contains the open set $V_{k}$. This means that $V_{k}\subset \pi^{-1}_{\gamma}(V_{\gamma}) $, so that after applying $\pi_{\gamma}$ on both sides we get $X_{\gamma}\subset V_{\gamma}$, contradicting the fact that $V_{\gamma}$ is a proper subet of $X_{\gamma}$.
