I can partially answer the second question. If $X$ is a compact Hausdorff space whose topology has a countable base at every point (e.g. $\omega_1$), then there are no bad sequences. Moreover the following holds:

If $F\subset C(X)$ is a family such that every countable subfamily has radius $\le 1$, then $r(F)\le 1$.

Define a function $S=S_F:X\to\widehat{\mathbb R}=[-\infty,+\infty]$ (the "essential supremum" of $F$) as follows: $S(x)$ is the maximum $t$ such that for every neighborhood $U$ of $x$ one has $\sup\{f(y):f\in F,y\in U\}\ge t$. Observe that $S$ is upper semi-continuous: for every $t\in\widehat{\mathbb R}$, the set $\{x\in X:S(x)\ge t\}$ is closed.

Define the essential infimum $I_F$ similarly, this function is lower semi-continuous.

For every $x\in X$ there is a countable family $G\subset F$ such that $S_G(x)=S_F(x)$ and $I_G(x)=I_F(x)$. Indeed, using countable base at $x$, one can realize $S(x)$ by a sequence $x_i:i\in\mathbb N$ converging to $x$ and functions $f_i\in F$ such that $f_i(x_i)\to S(x)$.

It follows that $S(x)\le I(x)+2$ for all $x\in X$. Indeed, take $G$ as above, it is contained in a $(1+\epsilon)$-ball centered at some $f\in C(X)$, then $S_G(x)\le f(x)+1+\epsilon$ and $I_G(x)\ge f(x)-1-\epsilon$.

Fix $\epsilon>0$ and let us prove that $F$ is contained in a $(1+\epsilon)$-ball.  For $x\in X$, define $C_x=\frac12(S(x)+I(x))$. Note that $S(x)\le C_x+1$ and $I(x)\ge C_x-1$. By semi-continuity, there is a neighborhood $U_x$ of $x$ such that $S(y)<C_x+1+\epsilon$ and $I(y)>C_x-1-\epsilon$ for all $y\in U_x$. Choose a finite subcovering $V_i=U_{x_i}:i\le N$.

On each neighborhood $V_i$ we have a constant function $f_i:=C_{x_i}$ which works as a center within this neighborhood. It suffices to construct a function $g\in C(X)$ such that for every $x\in X$, $g(x)$ is between the minimum and maximum of these partially defined constant functions at $x$. This is easy to do by induction in the number of sets in the covering. Suppose we have already defined $g=g_{n-1}$ that works on $\bigcup_{i<n} V_i$. By Urysohn's lemma there is $\phi:X\to[0,1]$ such that $\phi=0$ on $X\setminus V_n$ and $\phi=1$ on $X\setminus\bigcup_{i\ne n} V_i$. Then $g_n:=\phi f_n+(1-\phi)g_{n-1}$ works on $\bigcup_{i\le n}V_i$.