As mentioned in the comment, the key is to show $$\lim \sup_{k \to \infty} A_k = \{\lim \sup_{k \to \infty} |X_k - \mu| \geq \epsilon \}$$ You can do that in a couple steps, showing $\cup_{k \geq n} A_k = \{\omega \, : \,\, \sup_{k \geq n} |X_k(\omega) - \mu| \geq \epsilon \}$ holds for all $n$ using a double-inclusion argument.
Then the step of pulling the limit outside the probability is OK, as you said, from monotone limit properties of probability measures.
For the other direction, take $U_N = \{\lim \sup_{k \to \infty} |X_k - \mu| \geq 1/N \}$. Then you get $$P(U_N) = \lim_n P(\sup_{k \geq n} |X_k - \mu| \geq 1/N) = 0$$ for any $N$ by assumption.
$U_N \uparrow \{\lim \sup_{k \to \infty} |X_k - \mu| > 0 \} = U$, so $P(U) = \lim_N P(U_N) = 0$. That means
$$0 \leq \lim \inf_{k \to \infty} |X_k - \mu| \leq \lim \sup_{k \to \infty} |X_k - \mu| = 0$$ with probability one, so the limit exists and is zero.
About your question in italics, it is always true that for a real-valued random variable sequence $\{f_k\}$ you have for any $a \geq 0$ $$\lim \sup_{k \to \infty} \{|f_k| \geq a \} = \{\lim \sup_{k \to \infty} |f_k| \geq a \}$$
which is what you would prove in the step mentioned in the comment. But if you switch the inequality direction it doesn't need to hold.
There are some areas (such as computability theory) where it is a useful convention to say that
$$ \mathit{expr}_1 = \mathit{expr}_2 $$
means, "either both expressions are defined with the same value or neither of the expressions are defined".
But basic real analysis is not one of those areas. (And even in the areas where the convention is used, it is good form to explicitly say that you're using it before jumping right into equations).
The usual convention for $=$ is more or less that at best $\mathit{expr}_1 = \mathit{expr}_2$ is false when one or both of the expressions is undefined. At worst the expression is considered to be nonsense if we write something that depends on its truth value in a context where we're not sure both are defined.
Muddying the waters a bit further we have the notation $\lim_{x\to a}f(x)=+\infty$. The most common way to define this notation is that the entire combination of ink shapes "$\lim\cdots=+\infty$" is a single symbol and the result is not actually an equation where "$=$" has its usual meaning. The whole thing is just a conventional way to claim $f(x)$ fails in a particular way to have a finite limit.
However, it is also possible to declare that you're working in the extended real line, in which case $+\infty$ and $-\infty$ are honest points in your topological space that are fully capable of being values of a limit expression.
Depending on which of these conventions you use, the claim, for example
$$ \lim_{x\to 2} \frac{1}{(x-2)^2} = \lim_{x\to \pi/2}\tan^2 x $$
could either be considered to be perfectly good (and true), or to be nonsense.
Since you don't know which definitions your readers will favor, it is generally good form not to write something like this unless you also point out that you're considering limits in the extended real line.
If you want to consider both limits of the form $\lim f(x)=+\infty$ and of the form $\lim f(x)=\pm\infty$ (where the latter form is satisfied by, for example, $\lim_{x\to 0}1/x = \pm\infty$), then even the extended real line will not save you, because a single function can now have more than one true limit. Then you have to fall back to the "naive" way out, saying that $\lim f(x)=\cdots$ is not actually an equation, but an indivisible claim about the behavior of $f$, and that "$\lim f(x)$" is not to be understood as a (single-valued) expression.
Best Answer
You can already find a good discussion of this topic here and in many textbooks, blogs, etc.
An explicit example of events $A_n$ such that $\lim_{n\to\infty} \mathbb P(A_n)\not=\mathbb P(\limsup_{n\to\infty}A_n)$ is as follows. Let $\mathbb P$ be uniform measure on $[0,1]$. Let $B_{i,j}$ denote the interval $[\frac{i}{j},\frac{i+1}{j}]$ and let $(A_n)$ be the sequence $$ \begin{matrix} B_{0,1},\\ B_{0,2},&B_{1,2}\\ B_{0,3},&B_{1,3},&B_{2,3}\\ \end{matrix} $$ and so on, reading left to right then top to bottom. Then $\limsup_{n\to\infty}A_n=[0,1]$, whereas the probability of the sets steadily decreases to $0$: in row $k$, all sets have proability $1/k$.