Proving that this set is a limit in $\mathbf{Set}$ As a continuation of this question, I'm trying to prove that $$L=\{(x_I)_{I\in \mathbf I} : x_I\in D(I)\text{ for all } I\in\mathbf I \text { and } (Du)(x_I)=x_J \text{ for all } u:I\to J \text{ in } \mathbf I\}$$ together with the projections $p_J: (x_I)_{I\in \mathbf I}\mapsto x_J$ is a limit in $\mathbf {Set}$.
Let $(f_I: A\to D(I))_{I\in\mathbf I}$ be a cone on $D$. We need to define a map $\alpha: A\to L$ such that $p_I\circ \alpha=f_I$ for all $I$. (And show that such a map is unique.)
Note that for any $a\in A,$ $$(f_I(a))_{i\in \mathbf I}\in L.$$
So we can define $\alpha:A\to L$ by $$\alpha: a\mapsto (f_I(a))_{i\in \mathbf I}.$$ Then $p_I\circ \alpha=f_I$. 
Is this part correct?
It remains to prove that $\alpha$ is unique. On the intuitive level, it's pretty clear, but how to show this formally?
 A: For uniqueness of $\alpha$, we do the following.
Notational aside: I'm going to use capital $I$ for the index category and lower case $i$ for an object of $I$, rather than capital $I$s for both differing in boldness.
In this notation, we have the cone $\{f_i : A\to D(i) \}_{i\in I}$,
and we have the purported limit cone, $\{p_i : L\to D(i) \}_{i\in I}$.
Suppose $\beta : A\to L$ is a map satisfying $p_i\circ \beta = f_i$.
Then we know that $$\beta(a) = (b_i)_{i\in I},$$
for some $b_i\in D(i)$ by definition of $L$ as a subset of the product. 
Applying $p_i$ to both sides gives us 
$$b_i = p_i(\beta(a)) = f_i(a).$$
Thus $\beta(a)=\alpha(a)$ by definition of $\alpha$. Since this is true for any $a\in A$, we have $\beta=\alpha$, and thus $\alpha$ is unique.
Another way of thinking about this:
Since you've defined $L$ as a subset of the product, and the $p_i$s as the restrictions of the projection maps out of the product, since the constraint is that $\alpha$ be unique for the property that $p_i\circ\alpha = f_i$, this is already true by the corresponding uniqueness property of the product. I.e., since $L$ is a subset of the product, we can regard $\alpha$ as a map to the product. Then by the uniqueness property for the product, we know that if $p_i\circ\beta = f_i = p_i\circ \alpha$, then $\alpha = \beta$.
