Special fiber is geometrically connected if the generic fiber is under properness assumption? I am a beginner in algebraic geometry, and want to understand the following proof towards criterion of Néron–Ogg–Shafarevich for abelian varieties:

Why can Zariski's connectedness theorem imply fiber is geometric connected? I think this requires $f_*O_X=O_{Y}$ and follows by Stein decomposition. Is there another proof or generalization for this lemma?
 A: Let $Y = Spec(R)$. Let us consider the map induced by $f$
$f : \mathcal{O}_{Y} \rightarrow f_*\mathcal{O}_X$
Since $X$ is proper over $R$, we get that $M := f_*\mathcal{O}_X$ is a finite module over $R$. Note that $M$ is also a reduced $R-$algebra. Consider the natural base change map $\varphi^0(y)$ for $y \in Y$ not necessarily closed point
$\varphi^0(y) : R^0f_*\mathcal{O}_X = f_*\mathcal{O}_X \otimes k(y)\rightarrow H^0(X_y, \mathcal{O}_{X_y})$
where $X_y$ is the fiber over the point $y$. For $y = Spec(k)$, the map $Spec(k) \rightarrow Spec(R)$ is a flat map and hence we know this map to be an isomorphism by flat base change theorem. Also note that $X_y$ is geometrically connected by hypothesis and hence $H^0(X_k, \mathcal{O}_{X_k}) = k$. Thus we get that the $R-$module $M$ satisifes $M \otimes k \cong k$. More geometrically, this says $Spec(M) \otimes Spec(k) \cong Spec(k)$. That is the map $Spec(M) \rightarrow Spec(R)$ is a normalization map since both have same function fields. Since $R$ being a dvr is already a normal ring, hence $Spec(M) \xrightarrow{\sim} Spec(R)$ is an isomorphism. That is $M \cong R$. Thus one has 
$\mathcal{O}_Y \rightarrow f_*\mathcal{O}_X$
is an isomorphism. That is what was required.
All this is done in the following lemma. https://stacks.math.columbia.edu/tag/0AY8
