$X_n\to X$ a.s. $\iff \forall \varepsilon>0, \lim_{n\to \infty }\mathbb P\left\{\bigcup_{m\geq n}|X_m-X|>\varepsilon\right\}=0$ $X_n\to X$ a.s. $\iff \forall \varepsilon>0, \lim_{n\to \infty }\mathbb P\left\{\bigcup_{m\geq n}|X_m-X|>\varepsilon\right\}=0$
$\boxed{\Rightarrow}$ Suppose $X_n\to X$ a.s.. Then, if $\varepsilon>0$,
$$\mathbb P\left\{\bigcup_{m\geq 1}\bigcap_{n\geq m}|X_n-X|\leq \varepsilon\right\}=1\implies \mathbb P\left\{\bigcap_{m\geq 1}\bigcup_{n\geq m}|X_n-X|>\varepsilon\right\}=0.$$
Since $$\mathbb P\left\{\bigcap_{m\geq 1}\bigcup_{n\geq m}|X_m-X|>\varepsilon\right\}=\lim_{m\to \infty }\mathbb P\left\{\bigcup_{n\geq m}|X_n-X|>\varepsilon\right\},$$
we have the result.
$\boxed{\Leftarrow}$ Suppose $$\forall \varepsilon>0, \lim_{n\to \infty }\mathbb P\left\{\bigcup_{m\geq n}|X_m-X|>\varepsilon\right\}=0.$$
Let $\varepsilon>0$. Then, $$0=\lim_{n\to \infty }\mathbb P\left\{\bigcup_{m\geq n}|X_m-X|>\varepsilon\right\}=\mathbb P\{\limsup_{n\to \infty }|X_n-X|>\varepsilon\}.$$
Question 1 : Is the first part correct ?
Question 2: Does $\mathbb P\{\limsup_{n\to \infty }|X_n-X|>\varepsilon\}=0$ mean that $\mathbb P\{\limsup_{n\to \infty }|X_n-X|\leq \varepsilon\}=1$, or only $\mathbb P\{\liminf_{n\to \infty }|X_n-X|\leq \varepsilon\}=1$ ? If we have $\mathbb P\{\limsup_{n\to \infty }|X_n-X|\leq \varepsilon\}=1$, then it's done, but I'm not sure that it's true. Otherwise, how would you do ? 
 A: You are confusing two things; limsup of a sequence of random variables and limsup of a sequence of sets.

If $(A_n)_{n \in \mathbb{N}}$ is a sequence of sets, then $$\limsup_{n \to \infty} A_n := \bigcap_{n \in \mathbb{N}} \bigcup_{m \geq n} A_n$$ which implies that $$(\limsup_{n \to \infty} A_n)^c = \bigcup_{n \in \mathbb{N}} \bigcap_{m \geq n} A_n^c = \liminf_{n \to \infty} (A_n^c).$$ For $A_n := \{|X_n-X|>\epsilon\}$ this gives (in your framework) $$\mathbb{P}(\liminf_{n \to \infty} \{|X_n-X| \leq \epsilon\}) = 1-\mathbb{P}(\limsup_{n \to \infty} \{|X_n-X|>\epsilon\}) = 1.$$ This is equivalent to saying that with probability $1$ we have $|X_n(\omega) -X(\omega)| \leq \epsilon$ for $n \geq N(\omega)$ sufficiently large.

On the other hand, there is the limsup of a sequence of random variables $(Y_n)_n$
$$\limsup_{n \to \infty} Y_n(\omega)$$
which is defined as the largest accumulation point of the sequence $(Y_n(\omega))_{n \in \mathbb{N}}$ for $\omega \in \Omega$. Since
$$\{|Y| \leq \epsilon\}^c = \{|Y|>\epsilon\}$$
holds for any random variable $Y$, we have in particular
$$\mathbb{P}(\limsup_{n \to \infty} |Y_n| \leq \epsilon) = 1-\mathbb{P}(\limsup_{n \to \infty} |Y_n|>\epsilon).$$
Applying this for $Y_n := X_n-X$ gives (in your framework)
$$\mathbb{P}(\limsup_{n \to \infty} |X_n-X| \leq \epsilon)=1,$$
i.e. we  have with probability $1$
$$\limsup_{n \to \infty} |X_n-X| \leq \epsilon.$$
