Proposition 4.2.4 of Borceux's Handbook of Categorical Algebra Vol. 1 states that in a complete category, the intersection of every family of subobjects of a fixed object always exists.
The proof begins by taking a nonempty family of monomorphisms $s_i\colon S_i\to A$, for $i\in I$. By completeness, the limit $(L,(p_i)_{i\in I})$ of the diagram constituted by the various morphisms $s_i$ exists. He then says all the composites $s_i\circ p_i\colon L\to A$ are equal by definition of a limit.
This is my one stumbling block in the proof. I don't see what property of the limit makes these all equal. Are they all solutions to the same universal mapping problem?
Recall that a limit of a diagram $D : \mathcal I \to \mathcal C$ is in particular a cone over $D$, meaning an object $L$ together with a maps $(p_i)_{i\in\mathrm{Ob}(\mathcal I)}$ such that for any map $k:i \to j$ in $\mathcal I$ it holds that $D(k)\circ p_i = p_j$.
Here, starting from the family of monos $(s_i)_{i\in I}$, the category $\mathcal I$ is the category with objects:
The non identity morphisms of $\mathcal I$ are $k_i : i \to \star$ and there is exactly one such morphism for each $i\in I$. In other words it is the category obtained from the discrete set $I$ by formally adding a terminal object.
The diagram $D$ you are looking at is $i\mapsto S_i$, $\star \mapsto A$ and $k_i \mapsto s_i$. So given a cone $(L,(p_i)_{i\in \mathrm{Ob}(\mathcal I)}$), one has for each $i\in I$: $$s_i \circ p_i = D(k_i) \circ p_i = p_\star$$ So in particular all $s_i\circ p_i$ are equal. This is by definition of the limit being a cone over $D$.