About Rudin's theorem 7.25:
7.25 Theorem If $K$ is compact, if $f_n \in \mathscr{C}(K)$ for $n=1,2,3,\dots,$ and if $\{f_n\}$ is pointwise bounded and equicontinuous on $K$, then
- $\{f_n\}$ is uniformly bounded on $K$,
- $\{f_n\}$ contains a uniformly convergent subsequence.
The equicontinuity is needed for proving (1). But I don't see how it is needed for proving (2).
The proof is the following:
Proof
(1) Let $\varepsilon > 0$ be given and choose $\delta > 0$, in accordance with Definition 7.22, so that \begin{equation} |f_n(x) - f_n(y)| < \varepsilon \tag{44} \end{equation} for all $n$, provided that $d(x,y) < \delta$.
Since $K$ is compact, there are finitely many points $p_1, \dots, p_r$ in $K$ such that to every $x \in K$ corresponds at least one $p_i$ with $d(x, p_i) < \delta$. Since $\{f_n\}$ is pointwise bounded, there exist $M_i < \infty$ such that $|f_n(p_i)| < M_i$ for all $n$. If $M = \max(M_1, \dots, M_r)$, then $|f_n(x)| < M + \varepsilon$ for every $x \in K$. This proves (1).
(2) Let $E$ be a countable dense subset of $K$. (For the existence of such a set $E$, see Exercise 25, Chap. 2.) Theorem 7.23 shows that $\{f_n\}$ has a subsequence $\{f_{n_i}\}$ such that $\{f_{n_i}(x)\}$ converges for every $x \in E$.
Put $f_{n_i} = g_i$, to simplify the notation. We shall prove that $\{g_i\}$ converrges uniformly on $K$.
Let $\varepsilon > 0$, and pick $\delta > 0$ as in the beginning of this proof. Let $V(x,\delta)$ be the set of all $y \in K$ with $d(x,y) < \delta$. Since $E$ is dense in $K$, and $K$ is compact, there are finitely many points $x_1, \dots, x_m$ in $E$ such that $$ K \subset V(x_1, \delta) \cup \dots \cup V(x_m, \delta). \tag{45} \label{45} $$
Since $\{g_i(x)\}$ converges for every $x \in E$, there is an integer $N$ such that $$ |g_i(x_s) - g_j(x_s)| < \varepsilon \tag{46} \label{46} $$ whenever $i \geq N$, $j \geq N$, $1 \leq s \leq m$.
If $x \in K$, $(\ref{45})$ shows that $x \in V(x_s, \delta)$ for some $s$, so that $$ |g_i(x) - g_i(x_s) | < \varepsilon $$ for every $i$. If $i \geq N$ and $j \geq N$, it follows from $(\ref{46})$ that \begin{align} |g_i(x) - g_j(x)| &\leq |g_i(x) - g_i(x_s)| + |g_i(x_s) - g_j(x_s)| + |g_j(x_s) - g_j(x)| \\ &< 3\varepsilon. \end{align} This completes the proof.
So my question is: if I remove the equicontinuity from the assumptions, part (1) would not hold anymore, but would the part (2) of the theorem still hold?
No, it would not hold anymore. Think of $K=[0,1]$ and
$$ f_n(x)= \begin{cases} 1-nx, & x \in [0,1/n), \\0, & \text{else.} \end{cases}$$
The way $\delta$ is chosen in the proof above makes use of equicontinuity.