Show $\limsup\limits_{n\to \infty} X_n = \infty$ almost surely for $X_n$ uniform on $(0,n)$ and $(X_n)$ independent 
Let $(X_n)_{n \in \mathbb N}$ a sequence of independent random variables, where $X_n$ is uniformly continuous distributed across $(0,n)$. Show that $\limsup\limits_{n\to \infty} X_n = \infty$ almost surely. Hint: Use sets of the form $\{ X_n \geq b_n \}$ for a suitable $b_n \to \infty$.

I've tried this so far (unsure + might be incorrect):
I have to show almost sure convergence, so using this Definition
$$P(\limsup_{n\to \infty} X_n = X ) = 1$$
Lim Sup equals to the following (not sure about handling with infinity in the next step)
$$P(\inf_{k \in \mathbb n} \sup_{n\geq k}{X_n} = X ) = 1$$
If I now insert the $\{ X_n \geq b_n\}$: for $b_n := \sqrt n$ and use continuity from above from the lim sup, it equals(?) to
$$\lim_{n \to \infty} P(\{X_n \geq b_n \} ) = 1 \Rightarrow \lim_{n \to \infty} \frac{n-\sqrt n}{n} = 1 \Rightarrow \text{almost sure convergence}$$
Is this (somehow) correct?

If $b_n \to \infty$, then we find that
$$ \{ \limsup_{n\to\infty} X_n = \infty \} \supseteq \{ X_n \geq b_n \text{ infinitely often} \} = \bigcap_{N\geq 1} \bigcup_{n\geq N} \{ X_n \geq b_n \}. $$
So it suffices to show that $\mathbb{P}[X_n \geq b_n \text{ infinitely often}] = 1$ for some suitable choice of $(b_n)$. Now
\begin{align*}
1 - \mathbb{P} \left[ \bigcap_{N\geq 1} \bigcup_{n\geq N} \{ X_n \geq b_n \} \right]
&= \mathbb{P} \left[ \bigcup_{N\geq 1} \bigcap_{n\geq N} \{ X_n < b_n \} \right] \\
&\leq \sum_{N=1}^{\infty} \mathbb{P} \left[\bigcap_{n\geq N} \{ X_n < b_n \} \right] \\
&= \sum_{N=1}^{\infty} \prod_{n\geq N} \mathbb{P}[X_n < b_n ]
\end{align*}
Now if $(b_n)$ is chosen so that $b_n \to \infty$ and $\mathbb{P}[X_n \geq b_n] \to 1$, then you can easily check that this bound is exactly $0$, hence proving the desired claim.

Addendum. For a better resolution, let $M_n = \max\{X_1,\cdots,X_n\}$ and notice that
$$ \mathbb{P}[M_n \leq x]
= \prod_{i=1}^{n} \mathbb{P}[X_n \leq x]
= \prod_{i=1}^{n} \min\left\{ 1, \frac{x}{n} \right\}
= \dfrac{x^{n-\lfloor x\rfloor} \cdot\lfloor x\rfloor!}{n!}
\qquad \text{for } x \in (0, n). $$
Now if we plug $x = n - z\sqrt{n}$, then for each fixed $z \geq 0$ it is not hard to prove that
$$ \mathbb{P}[M_n \leq n - z\sqrt{n}] \xrightarrow[n\to\infty]{} e^{-z^2/2}. $$
