This is Exercise 7.1.D of Vakil's notes. A property $P$ being local on the source means that to check if a morphism $\pi: X \to Y$ has $P$, it suffices to check on any open cover $\{U_i\}$ of X.
I can't find an example showing why being an open embedding is not local on the source. In fact, let $\pi: (X, \mathscr{O}_X) \to (Y, \mathscr{O}_Y)$ is a morphism of schemes. Let $\{(U_i, \mathscr{O}_X|_{U_i})\}$ be an open cover of $X$. Suppose $\pi|_{U_i} = \pi_i$ is an open embedding for all $i$, then there are isomorphisms $(U_i, \mathscr{O}_X|_{U_i}) \xrightarrow[]{\rho_i}(V_i, \mathscr{O}_Y|_{V_i})$ where $V_i$ is an open subset of $Y$. Now these isomorphisms glue to an isomorphism $(X, \mathscr{O}_X) \to (\bigcup V_i, \mathscr{O}_Y|_{\bigcup V_i})$ where $\bigcup V_i$ is an open subset of $Y$. So $\pi$ is an open embedding.
What's wrong in my argument above? Can anyone give me a counterexample showing this is not open on the source? Thanks!
Consider a covering map. For example the obvious map $\mathbb{A}^1\sqcup \mathbb{A}^1 \to \mathbb{A}^1$.
The problem is that a map being a local homeomorphism doesn't guarantee that it is a global homeomorphism. In fact I can give you worse. Consider the map from the affine line with doubled origin to the affine line collapsing the double origin. This is locally an open embedding, but certainly not an isomorphism.
The problem with your argument is that even if the isomorphisms glue to a morphism, they don't necessarily glue to an isomorphism. This is only the case if their inverses also glue.