The morphism is monic iff the diagonal is an isomorphism

You don't have to check anything more; the argument is purely categorical.

To say that $f:X \to Y$ is monic is to say that if $g,h: Z \to X$ and $fg = fh$, then $g = h$. Now giving a pair of maps to $X$ so that $fg = fh$ is (by definition) the same as giving a map $Z \to X\times_Y X$, and saying that $g = h$ is the same as saying that this map factors through the diagonal.

So $f$ is monic precisely when any map $Z \to X\times_Y X$ factors through $\Delta: X \to X \times _Y X$. This automatically implies that $\Delta$ is an isomorphism, just by Yoneda. (If you were to unwrap the proof of Yoneda, you would find that you apply the preceding statement to the special case when $Z = X \times_Y X$ with the map to itself being the identity.)

The argument works in any category with fibre products.

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Regarding our last question, I'm not sure off the top of my head, never having thought about it.

"Also is there a monic morphism which is not both topolgically monic and sheaf epic?"

If you mean "sheaf epic" in the category of sheaves, then no, there are lots of them, already in the affine case ($=$ epimorphisms of commutative rings). See this seminar, especially "Monomorphismes de schémas noethériens" by Daniel Ferrand. See also MO/56591.

But I think that if $f : X \to Y$ is a monomorphism of schemes, then $f : X \to Y$ is injective (direct argument using spectra of fields, or EGA IV, 17.2.6) and $f^\# : \mathcal{O}_Y \to f_* \mathcal{O}_X$ is an epimorphism in the category of quasi-coherent $\mathcal{O}_Y$-algebras (let's assume $f$ qcqs so that $f_* \mathcal{O}_X$ is quasi-coherent by EGA I, 6.7.1). Namely, if $\alpha,\beta : f_* \mathcal{O}_X \to B$ are homomorphisms of quasi-coherent $\mathcal{O}_Y$-algebras, these correspond to two $Y$-morphisms $\mathrm{Spec}(B) \to X$, which therefore must be equal. [can someone check the details?]