$\newcommand{\Spec}{\mathrm{Spec}}$
Scholze---> Katz-Mazur: I really wouldn't stress too much about this, to be honest. Probably Scholze should say that $p$ is locally of finite presentation and/or $S$ is locally Noetherian. Since the moduli spaces of such objects constructed is locally Noetherian, you really have no harm restricting to such a thing. Then, proper implies finite type and since S is locally Noetherian this implies that $p$ is locally of finite presentation. And then, yes, we use
[Tag01V8][1] If it makes you feel any better, his ultimate goal with this paper, and subsequent ones (which, incidentally, my thesis is a generalization of one of these papers) is to work in the same realm as the work of Harris-Taylor. In Harris-Taylor's seminal book/paper where they prove local Langlands for $\mathrm{GL}_n(F)$ they explicitly restrict only the schemes which are locally Noetherian (as does Kottwitz, if I recall correctly, in his original paper "On the points of some Shimura varieties over finite fields).
Katz-Mazur ---> Scholze: A smooth proper connected curve over a field is automatically projective. We may assume we're over $\overline{k}$. Let $X$ be a smooth proper conneced curve. Let $U$ be an affine open subscheme. Then, by taking a projectivization of $U$ (i.e. locally closed immerse $U$ into some $\mathbb{P}^n$ and take closure) and normalizations you can find an $X'$ which is smooth and projective containing $U$. Then, you get a birational map $X\dashrightarrow X'$. One can then use the valuative criterion to deduce this is an isomorphism.
An elliptic curve is connected. Note then that if $X/k$ is finite type, connected, and $X(k)\ne \varnothing$ then $X$ is automatically geometrically connected. Since any idempotents in $\mathcal{O}(X_{\overline{k}})$ must show up at some finite extension, it suffices to show that $X_L$ is connected for every finite extension $L/k$. Note that since $\Spec(L)\to \Spec(k)$ is flat and finite then same is true for $X_L\to X$, and thus $X_L\to X$ is clopen. Thus, if $C$ is a connected component of $X_L$ it's clopen (since $X_L$ is Noetherian) and thus its image under $X_L\to X$ is clopen, and thus all of $X$. Suppose that there exists another connected component $C'$ of $X_L$. Then, by what we just said the image of $C$ and $C'$ both contain any $x\in X(k)$. Note though that if $\pi:X_L\to X$ is our projection, then $\pi^{-1}(x)$ can be identified set theoretically as $\Spec(L\otimes_k k)=\Spec(L)$ and co consists of one point. This means that $C$ and $C'$, since they both hit $x$, have an intersection point. This is a contradiction. So an elliptic curve, being connected and having $E(k)\ne \varnothing$, is automatically geometrically connected.
Here is a nice, basic fact.
Fact: Let $k$ be a field and $C$ a smooth geometrically integral projective curve over $k$. Then, for any open subset $U\subsetneq C$ is affine.
This is a somewhat easy exercise using Riemann--Roch to find for a given divisor $D\subseteq X$ (namely the complement of $U$) a large enough $m$ so that there is a rational function $f$ with poles precisely at the support of $mD$. You can see Lemma 11 here for the full proof (DISCLAIMER: this is my blog).
In particular, if $p$ is any closed point of $C$ (not necessarily a $k$-point) then $C-\{p\}$ is affine.
Question for you to ponder: Are the hypotheses I wrote above of smooth and goemetrically integral necessary?
Best Answer
This result is called Lüroth's theorem; see, e.g., this Math.SE question. For an elementary proof, you can look at Theorem 1.3 in this book by Ojanguren.
For more canonical references, you can look at Jacobson's Basic Algebra II, p. 522, or Hartshorne's Algebraic Geometry, Ex. IV.2.5.5.
Below is a version of Ojanguren's argument that I wrote down for a course taught by Mircea Mustaţă. See Lecture 1, Theorem 2.1 in his notes, or Theorem 1.6 in my LiveTeXed notes from the course.
Proof. We show that if $k \subseteq K \subseteq k(t)$ is a sequence of field extensions such that $\operatorname{trdeg}_k K = 1$ and $t$ is transcendental over $k$, then $K \simeq k(X)$ for a transcendental element $X$ over $k$. Since $\operatorname{trdeg}_k K = 1$, it is enough to find some $a \in K$ such that $K = k(a)$.
First, the second extension $K \subseteq k(t)$ must be algebraic, so $t$ is algebraic over $K$. Let $$ f(X) = X^n + a_1X^{n-1} + \cdots + a_n \in K[X] $$ be the minimal polynomial of $t$ over $K$. Since $t$ is transcendental over $k$, we cannot have all $a_i \in k$, and so there is some $i$ such that $a_i \in K \smallsetminus k$. We will show that in this case, $K = k(a_i)$. We know that the $a_i \in K \subseteq k(t)$, and so we may write $$ a_i = \frac{u(t)}{v(t)}, $$ where $u,v \in k[t]$ are relatively prime, and where at least one of them of positive degree by the assumption that $a_i \notin k$. Now consider the following polynomial: $$ F(X) = u(X) - a_iv(X) \in k(a_i)[X] \subseteq K[X]. $$ Since $F(t) = 0$, we have that $f(X) \mid F(X)$ in $K[X]$ by minimality of $f(X)$, and so \begin{equation} u(X) - a_iv(X) = f(X)g(X).\tag{1}\label{eq:0105star} \end{equation} where $g \in K[X]$. We then claim the following:
Claim. $g \in K$.
Showing the Claim would conclude the proof, for then we have a sequence of extensions $$ k \hookrightarrow k(a_i) \hookrightarrow K \hookrightarrow k(t) $$ where $t$ is the root of a degree $n$ polynomial in $k(a_i)[X]$, hence $[k(t) : K] \ge n = [k(t) : k(a_i)]$, which implies $K = k(a_i)$.
To prove the Claim, the first step is to get rid of all denominators: by multiplying \eqref{eq:0105star} by $v(t)$ and the product of all denominators in the coefficients of $f$ and $g$, we get a relation $$ c(t)\bigl(u(X)v(t) - v(X)u(t)\bigr) = f_1(t,X)g_1(t,X), $$ where $c(t) \in k[t]$ is nonzero and $f_1(t,X),g_1(t,X) \in k[t,X]$ are obtained from $f$ resp. $g$ by multiplication by an element in $k[t]$. Note that $k[t,X]$ is a UFD, hence every prime factor of $c(t)$ divides either $f_1(t,X)$ or $g_1(t,X)$. Thus, we can get rid of $c(t)$ by successively dividing by prime factors of $c(t)$ to get a relation \begin{equation} u(X)v(t) - v(X)u(t) = f_2(t,X)g_2(t,X),\tag{2}\label{eq:0105rel3} \end{equation} where now $f_2(t,X),g_2(t,X) \in k[t,X]$ are obtained from $f,g$ by multiplication by a nonzero element in $k(t)$. The trick is now to look at the degrees in $t$ on both sides. First, $$ \deg_t\bigl(u(X)v(t) - v(X)u(t)\bigr) \le \max\bigl\{\deg u(t),\deg v(t)\bigr\}. $$ Letting $f_2(t,X) = \gamma_0(t)X^n + \cdots + \gamma_n(t)$, we see that $$ \frac{\gamma_i(t)}{\gamma_0(t)} = a_i(t) = \frac{u(t)}{v(t)}, $$ where we recall that $u(t),v(t)$ were relatively prime. This implies that in fact, $$ \deg_t\bigl(f_2(t,X)\bigr) \ge \max\bigl\{\deg u(t),\deg v(t)\bigr\}. $$ By looking at degrees in $t$ on both sides of the relation \eqref{eq:0105rel3}, we have that $\deg_t(g_2(t,X)) = 0$, hence $g_2 \in k[X]$.
Now we show that $g_2 \in k$. For sake of contradiction, suppose that $g_2 \notin k$. Then, there is a root $\gamma \in \overline{k}$ such that $g_2(\gamma) = 0$, which implies that $u(\gamma)v(t) = v(\gamma)u(t)$ by \eqref{eq:0105rel3}. But since $u(t)$ and $v(t)$ are relatively prime, and are not both constants, we must have $u(\gamma) = 0 = v(\gamma)$. This contradicts that $u,v$ are relatively prime, and so $g_2 \in k$. Finally, since $g \in K[X]$ and $g = g_2 \cdot (\text{element of $k(t)$})$, we have that $g \in K$. $\blacksquare$