Let $I$ be some uncountable set, and let $\Bbb{R}^I$ denote the product of uncountably many copies of $\Bbb{R}$. Show that $\Bbb{R}^I$ is not metrizable.
I know that there is a solution which involves showing that $\Bbb{R}^I$ cannot be first countable, but I would like to try to show that this space isn't metrizable by showing the sequence lemma does not hold. Here's my idea:
If $c \in I$, let $1_c : I \to \Bbb{R}$ denote the indicator function on the singleton $\{c\}$, and let $S$ be the collection of all such functions. I'm pretty certain that $S$ is dense in $\Bbb{R}^I$, so given $\hat{2} : I \to \Bbb{R}$ defined by $\hat{2}(i) = 2$ for all $i \in I$, there is a sequence of indicator functions $1_{c_n}$ on singletons converging to $\hat{2}$. Since the projection $\pi_i$ onto the $i$-th factor is continuous, it follows that $\pi_i(1_{c_n}) \to \pi_i(\hat{2}) = \hat{2}(i) = 2$. This says that $2$ is a limit point of the set $\{\pi_i(1_{c_n}) \mid n \in \Bbb{N}\} = \{0,1\}$, which is absurd.
Something about what I did seems a bit off, but I can't identify it at the moment. E.g., the fact that I projected the sequence onto an arbitrary fact makes me nervous/suspicious.
EDIT:
In the mathematics chatroom, Ted Shifrin pointed out that $S$ is not in fact dense. For example, the basis element $B = \prod_{i \in I} U_i$ defined by $U_i = \Bbb{R}$ for all $i \neq c$ and $U_c = (2,3)$, where $c \in I$ is fixed. Since $(2,3)$ contains neither $0$ nor $1$, it's impossible to fit an indicator function in $B$.
So, it appears that the set $S$ may not be the best candidate in showing that the sequence lemma does not hold in $\Bbb{R}^I$.
Let $S$ be the set of points of $\mathbb{R}^I$ such that $\textrm{supp}(x) = |\{i \in I: x(i) \neq 0\}|$ is at most countable.
Any sequence in $S$ that converges, converges to a point of $S$. So if the sequence lemma held, $S$ would be closed. But $S$ is dense as can be easily checked (support set at most finite would have sufficed for that already).