The spectrum of a product of rings Let $A$ be the product of a family $(A_i)_{i\in I}$ of commutative rings, and  $c$ the canonical continuous map from the disjoint union $U$ of the spectra of the $A_i$ to the spectrum of $A$: 
$$
A:=\prod_{i\in I}\ A_i,\quad U:=\coprod_{i\in I}\ \text{Spec}(A_i),\quad c:U\to\text{Spec}(A).
$$

Is the image of $c$ dense?

Let $f$ be a continuous map from $U$ to a quasi-compact space $X$.

Is there a continuous map $F:\text{Spec}(A)\to X$ such that $F\circ c=f\ $?

[Diagram: 
$$
f:\coprod_{i\in I}\ \text{Spec}(A_i)\ \overset{c}{\to}\ \text{Spec}(A)\ \overset{F}{\to}\ X.]
$$
 A: The image of $c$ is dense if the $A_i$ are reduced. If $\mathrm{im}(c)$ is contained in some closed subset $V(f)$, then the latter contains the image of $V(\ker p_i)$ of the closed immersion $\mathrm{Spec}(A_i)\to \mathrm{Spec}(A)$,  where $p_i : A\to A_i$ is the canonical projection. So $f\in \sqrt{\ker p_i}$ and the $i$-component $f_i$ of $f$ is nilpotent, hence equal to $0$. This being true for all $i$, we have $f=0$. 
In general, if $f\in A$ is such that $f_i$ is nilpotent for all $i$, then the above reasonning shows that $V(f)$ contains the image of $c$. Now if there is such an $f$ which is not nilpotent, then $V(f)\ne \mathrm{Spec}A$ and the image of $c$ is not dense. For instance take $I=\mathbb N$, $A_i=\mathbb C[t_i]/(t_i^{i+1})$ and $f_i=\bar{t}_i$, then we have a counterexample.
The answer to the second part is negative without the quasi-compactness assumption on $X$. Just take $X=U$ and $f=\mathrm{Id}_U$. Then you would get a section $\mathrm{Spec}A\to U$ of $c$. 
Consider the example $I=\mathbb N$, $A_i=\mathbb F_{p_i}$ where $p_i$ is the $i$-th prime number (OK it is not the projection $p_i$ as above). Let $V_i=F^{-1}(\mathrm{Spec}(A_i))$. Then $\{ V_i\}_i$ is an open covering of $\mathrm{Spec}A$. For any principal open subset $D(f)$ contained in $V_i$, $p_i=0$ in $A_f$ because we have a homomorphism $A_i\to A_f$. Hence $p_if=0$ in $A$ (which is reduced). Projecting to $A_j$ for all $j\ne i$, we get $f_j=0$. Hence $D(f)$ is empty or equal to the singleton $x_i=\ker (A\to A_i)$. But it is well known by the theory of ultrafilters that $\mathrm{Spec}A$ is bigger than $\{ x_i\}_i$. 
