Show a scheme is integral if and only if it is irreducible and reduced. I'm trying to solve Exercise 5.2.F from Vakil's notes:

Show a scheme $X$ is integral if and only if it is irreducible and reduced.

Where we say $X$ is reduced (integral) if $\mathscr O_X(U)$ is reduced (an integral domain) for all open subsets $U$ of $X$.
Clearly, if $X$ is integral, then each $\mathscr{O}_X(U)$ is a domain, hence reduced, so $X$ is reduced. I'm not sure how to see that $X$ is irreducible. It's obvious for an affine scheme, since if $\mathscr{O}_X(X)=:A$ is a domain then $\text{Spec } A$ is irreducible. Am I able to use this to tackle the general case? Like if $X=\cup_i U_i$ with each $U_i$ affine open, does each $U_i$ being irreducible imply that $X$ is irreducible? This doesn't seem like the right way to approach it.
I'm not sure, I just feel stuck. Any hints would be greatly appreciated (I'd prefer that over someone just giving me the answer).

Hint for the forward direction: If $X$ is not irreducible, that means there are two disjoint nonempty open subsets $U,V\subset X$.  Consider $\mathscr{O}_X(U\cup V)$.
Hint for the reverse direction: If $X$ is irreducible and reduced, so is any open subscheme of $X$, so every nonempty affine open subscheme of $X$ is Spec of a domain.  To show $\mathscr{O}_X(U)$ is a domain for arbitrary $U$, you can show that the restriction $\mathscr{O}_X(U)\to\mathscr{O}_X(V)$ is injective for any nonempty affine open subset $V\subseteq U$.
A stronger hint for the reverse direction is hidden below:

 To show the restriction $\mathscr{O}_X(U)\to\mathscr{O}_X(V)$ is injective, suppose $f$ is in its kernel.  Then $f$ vanishes on $V$.  To show $f$ vanishes on $U$, it suffices to show $f$ vanishes on $W$ for every other affine open $W\subseteq U$.  Now use the fact that $W$ is Spec of a domain and $f$ vanishes on $V\cap W$.

