The theorem as stated is true (see my comment though): If 1), 2), 3), and 4) hold, then $f_n$ converges uniformly to $f$.
The answer to your third question is "yes". This is the contrapositive of the theorem, which is logically equivalent to the theorem.
I think the answer to your second question is no (specifically, 4) does not necessarily have to hold).
Other observations:
Remark 1:
The theorem is "tight" (?); that is, each of the hypotheses are needed to insure that the convergence is uniform.
To see this:
Take $K=[0,1]$.
The functions $f_n(x)=x^n$ give a sequence satisfying 1), 3), and 4), but not 2).
$f_n$ converges to $f=0$ pointwise but does not converge uniformly to $f$ on $K$.
Take $K=[0,\infty)$ and $f_n(x)=\begin{cases} 0,\; & 0\leq x\leq n\\x-n , &n< x\leq n+1\\ 1 ,& x>n+1\end{cases}$. This sequence satisfies 1), 2) and 4) but not 3).
$f_n$ converges to $f=0$ pointwise but does not converge uniformly to $f$ on $K$.
Take $K=[0,1]$ and let $f_n(x)$ be the function whose graph consists of the straight line segment from $(0,0)$ to $({1\over2n},1)$, the straight line segment from $({1\over2n},1)$ to $({1\over n}, 0)$, and the straight line segment from $({1\over n},0)$ to $(1,0)$. Then $f_n$ satisfies 1), 2), and 3), but not 4).
$f_n$ converges to $f=0$ pointwise but does not converge uniformly to $f$ on $K$.
Take $K=[0,1]$ and $f_n(x)=\begin{cases}1, & 0\le x\le 1-{1\over n}\cr 0,&1-{1\over n}< x<1\\1,\;&x=1\end{cases} $. This sequence satisfies 2), 3) and 4) but not 1).
$f_n$ converges to $f=1$ pointwise but does not converge uniformly to $f$ on $K$.
Remark 2:
If $(f_n)$ is uniformly convergent to $f$, you may not conclude that all four conditions hold. For example $f_n(x)=\begin{cases}1/n,& 0\leq x<1/2\\ -1/n& 1/2\leq x\leq1 \end{cases}$ converges uniformly to the zero function. But $(f_n)$ does not satisfy 1) or 4).
This shows that the converse of the theorem is false. It is not an "if and only if" theorem.
I will expand this post later (if that's allowed).
For $x\in \mathbb R,$ define $f_n(x) = n^2|x-1/n|.$ Then each $f_n$ is continuous everywhere, and $f_n \to \infty$ pointwise on $\mathbb R.$
Set $K = \{1/k: k \in \mathbb N \} \cup \{0\}.$ Then $K$ is compact. The only subsets of $K$ where $f_n \to \infty$ uniformly are the finite subsets. Proof: $f_n(1/n) = 0, n \in \mathbb N.$
As for Egorov's theorem, it certainly holds for pointwise convergence to $\infty.$ Just replace $f_n$ by $\arctan f_n.$ The latter sequence converges pointwise to the constant $\pi/2.$ Apply the usual Egorov to get uniform convergence to $\pi/2$ on a "large" subset of $K,$ then apply $\tan$ to get the desired conclusion.
Best Answer
It is one of the assumptions of Dini's theorem that the limit function $f$ is continuous. You can't use Dini's theorem to show that a monotone limit of continuous functions is continuous, because that is not true.
Counterexample: $$f_n(x)=\begin{cases} 0 & x\le 0 \\ x^n & 0<x<1\\ 1 & x\ge 1 \end{cases} $$