Surreal compactness In a comment here, Joel David Hamkins said:

...I think perhaps every set-sized open cover of a bounded interval in the surreals has a finite subcover, but there are proper class open covers with no set-sized subcover. (But I have to think more about this to be sure.) –  Joel David Hamkins 15 hours ago 

What are references (or proofs) for those two things?
I'll provide the second one, modulo the first one.
So: assume any set-size open cover of a closed interval admits a finite subcover.
Let $\omega$ be as usual, in particular
$\omega > n$ for all $n \in \mathbb N$.  Consider the closed interval
$$
[0,1] := \{x \in \mathbf{No}\;:\; 0 \le x \le 1\} .
$$
A proper-class open cover can be made up of all open intervals
$$
\left]a-\frac{1}{\omega}, a+\frac{1}{\omega}\right[
$$
where $a$ ranges over the proper class $[0,1]$.  But of course no finite set of
those covers $[0,1]$, since each of them can contain at most one rational number and $[0,1]$ has infinitely many rationals in it.
Thus we are left with:
Question:

Let $[0,1]$ be the closed interval in $\mathbf{No}$ defined above.  Is it true that any cover of $[0,1]$ by a set of open intervals admits a finite subcover?

 A: Suppose $\mathcal{A}$ is a set-size open cover of $[0,1]$ with open intervals. Without loss of generality, $\mathcal{A}$ is closed under overlapping unions (i.e. the union of any two elements of $\mathcal{A}$ that happens to be an open interval is again in $\mathcal{A}$). Let $B$ be the set of right endpoints $b$ of all intervals $(a,b) \in \mathcal{A}$ such that $a < 0$. We must show that $B$ contains a number bigger than $1$.
Let $\gamma$ be a sufficiently large ordinal, at least larger than all the birthdays of endpoints of intervals from $\mathcal{A}$ and larger than $\omega$. Let $c$ be the simplest number such that $b \leq c$ for every $b \in B$ and $c < d$ for every upper bound of $B$ born before time $\gamma$. If $B$ doesn't contain a number bigger than $1$, then certainly $c \leq 1$ since $1$ was born before time $\gamma$. So $c$ must be covered by an interval $(a,b)$ from $\mathcal{A}$. Since $a$ was born before time $\gamma$, it must be that $a < b'$ for some $b' \in B$. But then if $(a',b')$ is an interval from $\mathcal{A}$ with $a' < 0$ then $(a',b') \cup (a,b) = (a',b)$ shows that $b \in B$, which is impossible since $c < b$. From this contradiction, we conclude that $B$ must contain an element greater than $1$.
A: $\newcommand\No{\text{No}}$
Here is a proof of the second claim, that there is a proper class-sized open
cover of the surreal init interval with no set-sized subcover.
In particular, there is no finite subcover. (Note that your open cover does admit a set-sized subcover.)  The point is that $\No$ is not order-complete with respect to
proper class subsets. To see this, build two sequences
$$0=a_0<a_1<\cdots<a_\alpha <\cdots\cdots<b_\alpha<\cdots <b_1<b_0=1$$ of
length Ord, with no surreal number filling cut between them. Given
$a_\alpha< b_\alpha$, we can let $a_{\alpha+1}$ be the average of
$a_\alpha$ and $b_\alpha$, and $b_{\alpha+1}$ the average of
$a_{\alpha+1}$ and $b_\alpha$. At limits, the limits are not
equal, since every set-sized cut in $\No$ is filled with many
distinct surreal numbers, and so we can find $a_\lambda<b_\lambda$
in between, specifically $a_\lambda=\{a_\alpha\mid b_\alpha
\}_{\alpha<\lambda}$ and $b_\lambda=\{a_\lambda\mid b_\alpha\}_{\alpha<\lambda}$. Note that the birthdays of
these points $a_\alpha$ and $b_\alpha$ are strictly increasing as
$\alpha$ increases, and furthermore, the interval
$(a_\alpha,b_\alpha)$ contains no surreal numbers with earlier
birthdays, since $b_\alpha$ is the successor of $a_\alpha$ among
the surreals that are born on that day or earlier. It follows that
there can be no surreal number $z$ with $a_\alpha<z<b_\alpha$ for
all $\alpha$, because it would have to be born at some ordinal
stage $\alpha$, and then it would contradict the claim that
$(a_{\alpha+1},b_{\alpha+1})$ has no surreals with earlier
birthday.
Thus, let $U_\alpha=[0,a_\alpha)\cup(b_\alpha,1]$. This is a
proper-class open cover of $[0,1]$, but it has no set-sized
subcover, since any set of $U_\alpha$'s would miss out on the
later $a_\beta$'s.
Update. Regarding the open cover property, there are some subtle set/class issues here to be found. 
François has explained that every set-sized open cover of $[0,1]$
consisting of intervals admits a finite subcover, and it follows
immediately from this that every set-sized open cover consisting
of set-sized unions of intervals also admits a finite subcover,
since one may simply take the set of all intervals appearing in
any such union as another open cover and reduce to the finite
subcover.
But let me show that there are countably many open proper classes
$U_n$ for $n\in\omega$, such that $[0,1]\subset\bigcup_n U_n$, but
no finitely many $U_n$ cover $[0,1]$. In fact, we will arrange
that the $U_n$ are disjoint open classes, and so we will have an
open partition into open proper classes.
To see this, we use the fact that cuts of the type I constructed
above appear densely in the surreal numbers, that is, proper class
cuts $Z$ for which there is a lower sequence $a_\alpha$ converging
up to $Z$ and an upper sequence $b_\alpha$ converging down to $Z$,
with no surreal number filling this cut. Specifically, the reader
may construct countably many such cuts $Z_n$, each with a lower
sequence $a^n_\alpha$ converging up to it, and an upper sequence
$b^n_\alpha$ converging down to it, so that there is no surreal
number $z$ filling the cut, with $a^n_\alpha<z<b^n_\alpha$ for all
$\alpha$. We may construct these cuts so that $0<Z_0<Z_1<\cdots
<Z_n<\cdots<1$, meaning that the sequences $a^n_\alpha, b^n_\alpha$ are
all separated within the surreal unit interval $[0,1]$, and we may
assume furthermore that the upper and lower sequences for each
such cut do not overlap or interfere with one another.
Now, let $U_n=(Z_n,Z_{n+1})$, meaning the proper class open union $U_n=\bigcup_\alpha (b^n_\alpha,a^{n+1}_\alpha)$, which is a convex class in the surreals, but the endpoints are not realized in $\No$. Adding the bottom and top, let $U_{-1}=(-2,Z_0)\cup(\sup
Z_n,2)$, where by this latter interval, we mean the surreal
numbers above all the $b^n_0$'s and below $2$. This is an open
interval, since no increasing $\omega$-sequence converges.
Since the cuts $Z_n$ are not filled by any surreal number, it
follows that $[0,1]\subset \bigcup_n U_n$, and so we have a
countable open cover of the unit interval consisting of open
proper classes. But there is no finite subcover, since in fact
these open classes form a partition of $[0,1]$ into open classes.
So this is a sense in which the surreal numbers are connected,
with respect to separations into disjoint open sets, but it is
disconnected, if one allows the separating open classes to be
proper classes.
