Injectivity of base change morphism $(f_\ast \mathcal{O}_X)_s \otimes_{\mathcal{O}_{S,s}} k(s) \to H^0(X_s, \mathcal{O}_{X_s})$ Let $ f: X \to S$ be a proper flat, surjective morphism of schemes with $S$ locally Noetherian, and the geometric fibers of $f$ are reduced and connected. I'm trying to understand why $f_\ast \mathcal{O}_X = \mathcal{O}_S$.
The hypotheses of the question imply for any $s \in S$ that $k(s) = H^0(X_s, \mathcal{O}_{X_s})$.


Question 1: Why is the base change map $(f_\ast \mathcal{O}_X)_s \otimes_{\mathcal{O}_{S,s}} k(s) \to H^0(X_s, \mathcal{O}_{X_s}) = k(s)$ an isomorphism?


If I know this, I am in good shape for then cohomology and base change implies that $(f_\ast \mathcal{O}_X)_s$ is free, necessarily of rank $1$. Now to show that $\mathcal{O}_S \to f_\ast \mathcal{O}_X$ is an isomorphism it is enough for
$$ \mathcal{O}_{S,s} \otimes_{\mathcal{O}_{S,s}} k(s) \to (f_\ast \mathcal{O}_X)_s \otimes_{\mathcal{O}_{S,s}} k(s)$$
to be an isomorphism for all $s\in S$.


Question 2: For a fixed $s \in S$, this map is either zero or an isomorphism (because it is a linear map between the $k(s)$- vector spaces $k(s) \to k(s)$). Why is it not zero? 


 A: I find question 2 (or rather, the question of surjectivity that underlies it) easier than question 1.  
Namely, there is a tautological morphism $\mathcal O_S \to f_*\mathcal O_X$
(intuitively, this is pull-back of functions; formally, it is part of the very data defining the morphism of schemes $f:X \to S$).   Passing to the fibre at $s$, we obtain a morphism
$$\kappa(s) \to \kappa(s)\otimes (f_*\mathcal O_X)_s,$$ whose composite with the morphism
$$\kappa(s)\otimes (f_*\mathcal O_X)_s \to H^0(X_s, \mathcal O)$$ is the identity.  Thus this latter morphism is surjective.  
As for question 1, I don't see how to approach it directly.  Rather, I will prove the the base-change morphism is an isomorphism directly, using the theorem on formal functions as my main technical tool.  It lets me reduce to the case when $S = \mathrm{Spec} A$ with $A$ Artinian local, where I can argue directly.  Before I give the details, let me explain various possible reductions, and comment on their utility.
Reduction 1:  Since we have a natural morphism $\mathcal O_S \to f_*\mathcal O_X,$ it is a local problem to prove that it is an isomorphism, and so we may pass to the members of an open affine cover of $S$.  This lets us reduce to the case where $S = \operatorname{Spec} A$ is affine, and we have to show that $H^0(X,\mathcal O_X) = A$.  I don't know how to do this directly, though.
Reduction 2:  It's enough to check we get an isomorphism on stalks at points $s \in S$.  Also, the result on $H^0$ and flat base-change (see 
Hartshorne, Ch. III.9) applied to the flat morphism Spec $\mathcal O_{S,s}
\to S$, allows us to interpret the stalk at $s$ of $f_*\mathcal O_X$ as
$H^0(X \times_S \operatorname{Spec} \mathcal O_{S,s})$.  Thus we reduce to the situation
of Reduction 1, but with $A$ furthmore assumed to be local.  Again, though, I don't know how to directly argue from this point.
Reduction 3: If $\widehat{A}$ denotes the completion of the local ring $A$, then $A \to \widehat{A}$ is faithfully flat.  Base-changing the situation of Reduction 2 over this map, we reduce to the case where $A$ is complete local.  Once again, I'm not sure how to argue directly from this set-up.
Reduction 4:  We now use the assumption that $f$ is proper (or assume it is projective, if Hartshorne is your reference) and invoke the theorem on formal functions.  A consideration of that result shows that to prove the result when $A$ is complete local Noetherian, it in fact suffices to prove it in the case when $A$ is Artinian local. This is what I'll now do.
Thus I assume $X$ is proper and flat over the Artinian local ring $A$, and 
that its special fibre is geometrically connected and geometrically reduced.  
If $M$ is a finitely generated $A$-module, then there is a corresponding sheaf $M\otimes_A \mathcal O_X$ on $X$ (if you like, it is the pull-back via $f$ of the coherent sheaf on Spec $A$ attached to $M$ --- which Hartshorne denotes by $\widetilde{M}$, I think, but which I prefer to  describe as $M \otimes_A \mathcal O_S$, where $S = $ Spec $A$), and there is a natural map 
$$M = H^0(S, M\otimes_A \mathcal O_S) \to H^0(X,M\otimes_A \mathcal O_X).$$
I will show that this map is an isomorphism.  Taking $M = A$ will give the desired isomorphism $A \cong H^0(X,\mathcal O_X)$.
The proof is by induction on the length of $M$, with the case when $M$ has length one reducing to the fact that $\kappa(s) = H^0(X_s,\mathcal O_{X_s}).$ (As you note in the OP, this is where the assumption on the special fibre comes in.)  For the inductive step, consider a short exact sequence
$$0 \to M' \to M \to M'' \to 0$$ of $A$-modules.
By flatness of $X$ over $A$ (and this is the place that we use flatness) we obtain a short exact sequence of sheaves 
$0 \to M'\otimes_A \mathcal O_X \to M\otimes_A \mathcal O_X \to
M''\otimes_A \mathcal O_X \to 0,$
and hence an exact sequence
$$0 \to H^0(X,M'\otimes_A \mathcal O_X) \to H^0(X,M\otimes_A \mathcal O_X) \to H^0(X,M''\otimes_A \mathcal O_X).$$  The natural morphism from the exact sequence of $A$-modules to this exact sequence of global sections is an isomorphism on the outer terms --- by induction --- and so a five-lemma style argument shows it induces an isomorphism 
$$M \cong H^0(X, M\otimes_A \mathcal O_X).$$
This completes the proof.

I will make just a few comments on the intuitions behind the argument, and how I came up with it, although to a large part it is just a matter of experience and training.
One thing is that reductions steps 1, 2, and 3 are completely routine, and you should always have them in mind when thinking about this kind of question dealing with push forward of sheaves.
When a morphism is proper, the one big additional tool you have is the theorem on formal functions, and it frequently lets you perform reduction step 4, i.e. reduce to the case of an Artinian local base.  
In your question, the case of an Artinian local base seems natural to think about anyway, since working with a non-reduced base exerts maximal tension on the assumption that the fibre is reduced.   For myself, I thought about the case where the base is the dual numbers Spec $k[\epsilon]$ first, where one gets an easy argument based on the short exact sequence coming from multiplication by $\epsilon$.  The argument in the general Artinian case is an easy extension of that special case.  Note also that, for an Artinian local ring, flatness is much more tractable than in general, being closely related to concrete notions like length, so you can expect to get more purchase from it in this case than in more general contexts.

I don't know if there's a way to prove the result using your approach via base-change results that avoids the theorem on formal functions.  But as I wrote above, the theorem on formal functions is a very natural tool to use for this kind of question.
