There is likely a proof somewhere on this site but I could not find it. Here I give a quick proof of my comment (since I originally mis-stated the result by forgetting the "lower bounded" restriction):
Let $\{X_i\}_{i=1}^{\infty}$ be a sequence of random variables, not necessarily identically distributed and not necessarily independent, that satisfy:
i) $E[X_i]=m_i$, where $m_i \in \mathbb{R}$ for all $i\in\{1, 2, 3, ...\}$.
ii) There is a constant $\sigma^2_{bound}$ such that $Var(X_i) \leq \sigma^2_{bound}$ for all $i \in \{1, 2, 3, ...\}$.
iii) The variables are pairwise uncorrelated, so $E[(X_i-m_i)(X_j-m_j)]=0$ for all $i \neq j$.
iv) There is a value $b \in \mathbb{R}$ such that, with prob 1, $X_i-m_i\geq b$ for all $i \in \{1, 2, 3, ...\}$.
Define $L_n = \frac{1}{n}\sum_{i=1}^n (X_i-m_i)$. Then $L_n\rightarrow 0$ with prob 1.
Proof: Since the variables are pairwise uncorrelated with bounded variance, we easily find for all $n$:
$$ E[L_n^2] = \frac{1}{n^2}\sum_{i=1}^n \sigma_i^2 \leq \frac{\sigma_{bound}^2}{n} $$
Fix $\epsilon>0$. It follows that:
$$ P[|L_n|>\epsilon] = P[L_n^2 \leq \epsilon^2] \leq \frac{E[L_n^2]}{\epsilon^2} \leq \frac{\sigma_{bound}^2}{n\epsilon^2} $$
Hence:
$$ \sum_{n=1}^{\infty} P[|L_{n^2}|>\epsilon] \leq \sum_{n=1}^{\infty}\frac{\sigma_{bound}^2}{n^2\epsilon^2} < \infty $$
and so $L_{n^2}\rightarrow 0$ with probability 1 by the Borel-Cantelli Lemma. That is, the $L_n$ values converge over the sparse subsequence $n\in\{1, 4, 9, 16, ...\}$.
Since $L_n \geq b$ for all $n$ and $L_{n^2}\rightarrow 0$ with probability 1, it can be shown that $L_n\rightarrow 0$ with probability 1. $\Box$
The lower bounded condition is typically treated by writing $X_n = X_n^+ - X_n^-$ where $X_n^+$ and $X_n^-$ are nonnegative and defined $X_n^+=\max[X_n,0]$, $X_n^-=-\min[X_n,0]$. If $X_n$ and $X_i$ are independent, then $X_n^+$ and $X_i^+$ are also independent. So the lower bounded condition can be removed for the case when variables are independent. However, if $X_n$ and $X_i$ are uncorrelated, that does not mean $X_n^+$ and $X_i^+$ are uncorrelated. So it is not clear to me if the lower-bounded condition can be removed when "independence" is replaced by the weaker condition "pairwise uncorrelated."
Regarding Question 1: Yes, the probability of the event $\{ |\frac{X_1+\cdots + X_n}{n} - \mu | > \epsilon \}$ might be nonzero for any fixed $n$. Think of the following example: $X_1,...,X_n \stackrel{iid}{\sim} N(0,1)$. Then the probability of interest can be calculated exactly, since $\frac{X_1+\cdots + X_n}{n} \sim N(0, 1/n)$, and $P\{ |\frac{X_1+\cdots + X_n}{n} | > \epsilon \}= 2\Phi(-\epsilon \sqrt n)$, which converges to zero as $n$ goes to infinity, but is strictly positive for all $n$. Depending on what you assume about the random variables, you might be able to get a better description of how fast this convergence takes place. For example, if the $X_i's$ have a finite variance, then $P\{ |\frac{X_1+\cdots + X_n}{n} -\mu | > \epsilon \} \le Var(X_i)/(n\epsilon)$, which gives you an idea of how fast this probability decreases. This is called Chebyshev's inequality, and it can be improved, AKA the speed this upper bound approaches zero as a function of $n$ increased, if the $X_i's$ have higher order moments, are Gaussian (or subgaussian), etc..
Regarding Question 2: The same example implies $P\{ |\frac{X_1+\cdots + X_n}{n} - \mu | < \epsilon \}$ may not ever be equal to 1 for any $n$. That is not really what the strong law is guaranteeing. To really understand the strong law, you have to remember that random variables are just (measurable) functions mapping a probability space $(\Omega, \mathcal{F},P)$ to the real numbers. For every point of the sample space $\omega \in \Omega$, $|\frac{X_1(\omega)+\cdots + X_n(\omega)}{n} - \mu |$ defines a sequence of real numbers indexed by $n$. The strong law states that if you look at the subset $A \subset \Omega$ where $\omega \in A \iff |\frac{X_1(\omega)+\cdots + X_n(\omega)}{n} - \mu | \to 0, \;\; as \;\; n\to \infty$, then $P(A)=1$. In other words, the set of exceptional points in $\Omega$ where the functions $\frac{X_1(\omega)+\cdots + X_n(\omega)}{n}$ does not converge to $\mu$ must have probability zero. It is a worthwhile exercise to think about, and prove rigorously, why this is a stronger statement than the weak law, although normally this would not be taught until you are taking a course on measure based probability. See this post if your are interested Strong law of large numbers implies weak law
Best Answer
If any sequence has the $\limsup$ of its absolute values equal to 0, then the sequence converges to 0. Simply the $\liminf$ is sandwiched between 0 and the $\limsup$.
I have not read the book you mention. Unless the author explicitly explains, I can only speculate: maybe the $\limsup$ proof of the fact is more direct(?). It is a "weaker" result at face value. However, from the assertion stated above, it is in fact completely equivalent to the "$\lim$" statement.