Uniform versus product topologies on $[0,1]^\mathbb{N}$, and their Borel $\sigma$-algebras Let $\tau_U$ and $\tau_P$ be the uniform (i.e. sup-metric) and product topologies on $[0,1]^\mathbb{N}$, respectively.  Clearly, these topologies are not the same ($\tau_P$ is separable and $\tau_U$ is not, for instance).  However, a typical basic open set in $\tau_U$,
$V_{x,\epsilon} = \{ y : \forall n\in\mathbb{N}\;\; |x_n - y_n| < \epsilon \}$
is easily seen to be a $G_\delta$ set in $\tau_P$.
Question 1: Is every open set in $\tau_U$ in the $\sigma$-algebra generated by $\tau_P$?
My guess is "no", although I haven't been able to come up with a counterexample so far.  Assuming I'm right, though, I have the following weaker question;
Question 2: Does every open set in $\tau_U$ differ from an open set in $\tau_P$ by a meager set?  I.e., does every open set in $\tau_U$ have the Baire Property?
 A: The answer to question 1 is no.  As shown in this answer, the $\sigma$-algebra generated by any countable family of sets (such as the Borel $\sigma$-algebra of a separable metrizable space, which $\tau_P$ is) has cardinality at most $\mathfrak{c}$.  
On the other hand, $\tau_U$ contains $2^\mathfrak{c}$ distinct open sets.  For any $\mathcal{A} \subset 2^\mathbb{N}$, let
$$U_\mathcal{A} = \bigcup_{A \in \mathcal{A}} B(1_A, 1/2).$$
Clearly $U_\mathcal{A}$ is open.  Since for any $A \subset \mathbb{N}$ we have $1_A \in U_\mathcal{A}$ iff $A \in \mathcal{A}$, we have $U_\mathcal{A} = U_{\mathcal{A}'}$ iff $\mathcal{A} = \mathcal{A}'$.  Hence $\{U_\mathcal{A} : \mathcal{A} \in 2^\mathbb{N}\}$ are $2^{2^\mathbb{N}}$ distinct open sets in $\tau_U$.
A: Edit: After thinking about this a little longer, it seems the proof is much simpler.
Note that $[0,1]^\mathbb{N}$ contains the Cantor space $\{0,1\}^\mathbb{N}$.  The Cantor space is closed in $[0,1]^\mathbb{N}$ in both $\tau_U$ and $\tau_P$, and the restriction of $\tau_P$ to $\{0,1\}^\mathbb{N}$ produces the usual topology on the Cantor space whereas the restriction of $\tau_U$ gives the discrete topology.  Hence any subset of the Cantor space which doesn't have the Baire Property is already closed as a subset of $[0,1]^\mathbb{N}$ with $\tau_U$, and cannot have the Baire Property as a subset of $[0,1]^\mathbb{N}$ with $\tau_P$.
I'm leaving the previous argument below because I think it's interesting.

Nate's suggestion in his comment above works, with a little tinkering, to show that the answer to question 2 is no.
First, notice that we may replace $[0,1]^\mathbb{N}$ with $\mathbb{T}^\mathbb{N}$, where $\mathbb{T}$ is the unit circle. (Consider the quotient map $[0,1]\to \mathbb{T}$; all of the relevant topological notions are preserved.)
Now fix a function $\phi : \mathbb{T}^\mathbb{N}\to \mathbb{T}$ which is multiplicative, continuous with respect to the uniform topology on $\mathbb{T}^\mathbb{N}$, and takes convergent sequences to their limits.  (We won't need shift-invariance; any nonprincipal ultrafilter limit will do.)  Divide $\mathbb{T}$ into three closed arcs $A_0$, $A_1$, and $A_2$ corresponding to the intervals $[0,1/3]$, $[1/3,2/3]$, and $[2/3,1]$, and let $E_k = \phi^{-1}(A_k)$.  Note that each $E_k$ is uniformly closed, $\mathbb{T}^\mathbb{N} = E_0\cup E_1\cup E_2$, and
$$ E_0^2 = E_0\cup E_1,\quad E_1^2 = E_0\cup E_2,\quad E_2^2 = E_1\cup E_2 $$
One of the $E_k$'s must be nonmeager.  Suppose for a contradiction that it also has the Baire Property; then it's comeager in some nonempty (product-)open set $U\subseteq\mathbb{T}^\mathbb{N}$.  By Pettis's lemma (see Nate's reference above, or Theorem 9.9 in Kechris's Classical Descriptive Set Theory), $U^2\subseteq E_k^2$.  Then the image of $U^2$ under $\phi$ is disjoint from the interior of some $A_j$.  It's easy to see that this is a contradiction, since any (product-)open set in $\mathbb{T}^\mathbb{N}$ contains sequences which converge to any given limit in $\mathbb{T}$.
