I need to prove that category $\mathrm{Met}$ of metric spaces and continuous maps doesn't possess uncountable product of non-one point spaces.
Definition. A pair $(X,\{\pi_\nu:\nu\in\Lambda\})$ where $X\in\mathrm{Ob(Met)}$, $\pi_\nu\in \mathrm{Hom_{Met}}(X,X_\nu)$ is called a product of family of metric spaces $\{X_\nu:\nu\in\Lambda\}$ if for each $Y\in\mathrm{Ob(Met)}$ and $\{\varphi_\nu:\nu\in\Lambda\}$ where $\varphi_\nu\in \mathrm{Hom_{Met}}(Y,X_\nu)$ there exist unique $\varphi\in\mathrm{Hom_{Met}}(Y,X)$ such that $\varphi_\nu=\pi_\nu\varphi$.
My question. Assume that for all $\nu\in\Lambda$ the space $X_\nu$ contains at least two points. How one can prove that for such family of metric spaces their product doesn't exist in $\mathrm{Met}$?
My attempt We can consider arbitrary set $Y$ with discrete metric then every set theoreitc map $\varphi_\nu$, $\nu\in\Lambda$ will be continuous. Hence we see that if product $X$ exist, then it has to be set theoretic product of sets $\{X_\nu:\nu\in\Lambda\}$. On the other hand, we can consider this imaginary product as object of category topological spaces $\mathrm{Top}$. If we could extend universal property of $X$ to the cases when $Y$ is just topological space, then we would have that $X$ is a Tychonoff product. Since $\Lambda$ is uncountable and spaces $\{X_\nu:\nu\in\Lambda\}$ are not singletons, then topology of $X$ is not first countable and hence not metriazible. But this not a proof, since we can not extend universal property.
Could you give me a hint, or give me some related references for my question?