Are uncountable products of $\mathbb{R}$ always non-sequential?
I think it might depend on how weak you allow your set theory to be. If you consider set theory to be ZFC (and its extensions), then user642796 is correct that uncountable products of $\mathbb{R}$ are always non-sequential. If you allow set theory to be weaker than ZFC (in particular by removing the Axiom of Choice), then it's possible that Henno Brandsma is correct. I'll speak a bit more (but significantly less authoritatively) about this after showing the ZFC case.
To show (in ZFC) that $\mathbb{R}^\kappa$ is non-sequential whenever $\kappa$ is an uncountable cardinal, consider the subset $$A = \{ x = (x_\xi)_{\xi < \kappa} \in \mathbb{R}^\kappa : | \{ \xi < \kappa : x_\xi \neq 0 \} | \leq \aleph_0 \}.$$ This can easily be shown to be a non-closed subset of $\mathbb{R}^\kappa$ (it's actually a dense subset of $\mathbb{R}^\kappa$ which is clearly not all of $\mathbb{R}^\kappa$). We now show that no sequence in $A$ can converge to a point outside of $A$.
Let $( a_n = ( a^{(n)}_\xi )_{\xi < \kappa} )_{n \in \mathbb{N}}$ be a convergent sequence of points in $A$ with limit $b = ( b_\xi )_{\xi < \kappa}$. Note that as for each $n \in \mathbb{N}$ the set $\{ \xi < \kappa : a^{(n)}_\xi \neq 0 \}$ is countable, then so is $K = \bigcup_{n \in \mathbb{N}} \{ \xi < \kappa : a^{(n)}_\xi \neq 0 \}$. If $b \notin A$, then as $\{ \xi < \kappa : b_\xi \neq 0 \}$ is uncountable, there must be a $\xi \in \kappa \setminus K$ such that $b_\xi \neq 0$. As $a_n \rightarrow b$ it must be that $a^{(n)}_\xi \rightarrow b_\xi$ (projection mappings are continuous and so preserve sequential limits). But $a^{(n)}_\xi = 0$ for all $n$, and $b_\xi \neq 0$, which is absurd! Therefore it must be that $b \in A$, as required.
An observant reader will notice that I make an audacious use of the Axiom of Countable Choice (ACω) to conclude that $K$ is countable. (I think the statement that "countable unions of countable sets are countable" is strictly weaker than ACω, but ACω seems to be the weakest "quotable" axiom implying it.) It might be possible that uncountable products of $\mathbb{R}$ are sequential in some models of ZF+¬ACω... perhaps $\mathbb{R}^{\omega_1}$ is sequential in a model where $\omega_1$ is a countable union of countable sets.
I am also implicitly using the Axiom of Choice to reduce "all uncountable products of $\mathbb{R}$" to "every $\mathbb{R}^\kappa$ for an uncountable cardinal $\kappa$". If ACω fails, then you could have Dedekind finite but infinite sets (which are "uncountable", but are not equipotent to any ordinal). So this is also something to keep in mind. Mind you, considering $\mathbb{R}^I$ for an uncountable set $I$, we don't really run into any problems until trying to show that $\{ i \in I : b_i \neq 0 \} \setminus \bigcup_{n \in \mathbb{N}} \{ i \in I : a^{(n)}_i \neq 0 \}$ is nonempty.