If $F$ a sheaf and $S\subset F$ a subfunctor, then $S$ is a subsheaf if and only if... This is Proposition 1 from Maclane & Moerdijk's Sheaves in Geometry and Logic, part II, section 1.

Proposition 1. Let $F$ be a sheaf on $X$ and $S\subset F$ a subfunctor. $S$ is a subsheaf if and only if, for every open set $U$ on $X$, every element $f\in F(U)$, and every open covering $U=\bigcup U_i$, one has $f\in S(U)$ if and only if $f|_{U_i}\in S(U_i)$ for all i.

Proof: The stated condition is clearly necessary for $S$ to be a sheaf. Conversely, consider the commutative diagram 
$$\require{AMScd}
\begin{CD}
S(U) @>>> \prod S(U_i) @>>> \prod S(U_i\cap U_j)\\
@VVV & @VVV & @VVV\\\
F(U) @>>> \prod F(U_i) @>>> \prod F(U_i\cap U_j)
\end{CD}$$
The bottom row an equalizer. The last condition of the proposition states that the left hand square is a pullback. I have to prove that the top row is an equalizer. 
Can someone help me complete/formalize the proof?

Having reduced the problem this far, the argument no longer needs any information about $S$ or $F$: such a diagram's upper left corner is an equalizer in any category. That said, I'll write a proof using the names you've provided.
Suppose $f$ is equalized by the two top-right maps. Then the composition of $f$ with the middle vertical map is equalized by the two bottom-right maps, so factors through $FU$. Now we have maps to $FU$ and $\prod SU_i$ which agree at $\prod FU_i$, so they simultaneously factor through $SU$, and in particular $f$ does, so $SU\to \prod SU_i$ is an equalizer.
