I'm trying to understand the following proof:
Proposition. Let $(X,\mathcal{O}_X)$ be a ringed space. Then the category of $\mathcal{O}_X$-modules has enough injectives.
Proof. Let $\mathcal{F}$ be a sheaf. For every $x \in X$, embed the stalk into an injective $\mathcal{O}_{X,x}$-modules, $\mathcal{F}_x \to I_x$. Let $j$ denote the inclusion of $\{x\}$ into $X$ and let $\mathcal{I} = \prod_{x \in X}j_*(I_x).$ (...)
I cut out the rest because I cannot decipher what it means by $j_*(I_x)$. I know that $j_*$ means the pushforward induced by $j$, but for that to make sense, doesn't $I_x$ have to be a sheaf? In this state, $I_x$ is just an $\mathcal{O}_{X,x}$-module and I have no idea how to 'canonically' make sense of this as a sheaf on the singleton set $\{x\}$.
You must regard $I_x$ as a sheaf on the one point space. What have you tried? the topology of the singleton only has two opens, the empty set and the whole singleton. The sheaf condition fixes $I_x(\emptyset)$ (why?). Now, what should $I_x(\{x\})$ be? Define this presheaf and prove it is a sheaf.
Once you do this, you can easily generalize it: a sheaf of sets on $\{x\}$ is just a set, a sheaf of rings is just a ring, a sheaf of groups is just a group, etc.