Maybe it's easier to argue directly, like this:
Set $g_k=\inf_{n\ge k}x_n\ \text{and}\ h_k=\sup_{n\ge k}x_n\tag1$ Now since
$\liminf (x_n):=\underset {k\to \infty}\lim g_k,\ \limsup (x_n):=\underset{k \to \infty}\lim h_k\ \text{and}\ \ g_k\le x_k\le h_k,\tag2$
if $\ \liminf (x_n)=\limsup (x_n),\ \tag3 $
then $(x_n)$ converges by the squeeze theorem.
On the other hand, if $(x_n)\to L$, then there is an integer $N$ such that $L-\epsilon< x_n< L+\epsilon$ whenever $n\ge N$. Then, by definition of $(g_k)$ and $(h_k),$ and because $(g_k)$ is increasing and $(h_k)$ is decreasing, we have, for $n>N,$
$L-\epsilon\le g_N\le g_n\le h_n\le h_N\le L+\epsilon\tag 4$
and this implies that $(g_n)$ and $(h_n)$ converge to $L.$
It semms alright, up to some typos. But it would be simpler to say that $1_{\{X_n>\alpha\}}$ converges to $1$ almost surely, and so does $\mathbb E[1_{\{X_n>\alpha\}}]=\mathbb P(X_n>\alpha)$ by the dominated convergence theorem.
EDIT: apparently I was a bit fast, so let me give more details.
We suppose that $\mathbb P(X_n\underset{n\to+\infty}{\longrightarrow}\beta)=1$, or in other words, $X_n$ converges almost surely to $\beta$ as $n\to+\infty$. Let $\alpha<\beta$. We want to show that $\mathbb P(X_n>\alpha)\underset{n\to+\infty}{\longrightarrow}1$.
To do so, we can first observe that $\mathbb P(X_n>\alpha)=\mathbb E[1_{\{X_n>\alpha\}}]$. The interest of this trick is that now that it is formulated as the convergence of an expected value, we can use all the theorems we know about convergences of expected values, such that dominated convergence, monotonous convergence, etc. As a reminder, the dominated convergence theorem states that if $(Y_n)_{n\in\mathbb N}$ is a sequence of real-valued random variables such that:
(i) $Y_n$ converges almost surely to some random variable $Y$ ;
(ii) There exists an integrable random variable $Z$ such that for all $n\in\mathbb N$, $\vert Y_n\vert\le Z$ (almost surely),
then $\mathbb E[Y_n]\underset{n\to+\infty}{\longrightarrow}\mathbb E[Y]$ (and we even have that $Y_n$ converges in $L^1$ to $Y$, but the convergence of the expected values is sufficient for our purpose, as it is often the case).
Now just apply the latter with $Y_n=1_{\{X_n>\alpha\}}$. For almost all $\omega\in\Omega$, $X_n(\omega)$ converges to $\beta$ as $n\to+\infty$, which implies that for $n$ large enough, $X_n>\alpha$, or equivalently, $Y_n(\omega)=1$. Therefore, $Y_n$ converges almost surely to $1$. Moreover, $Y_n$ is clearly dominated by $1$ (so we take $Z=1$, see item (ii)). Therefore, by the dominated convergence theorem,
$$
\mathbb P(X_n>\alpha)=\mathbb E[1_{\{X_n>\alpha\}}]=\mathbb E[Y_n]\underset{n\to+\infty}{\longrightarrow}\mathbb E[Y]=1.
$$
Best Answer
For the sake of curiosity, you can prove a more general result which states:
Proof
According to the definition of limit, for every $\varepsilon > 0$ there is a natural number $N_{1}\geq 1$ such that \begin{align*} n\geq N_{1} \Rightarrow |a_{n} - a|\leq \varepsilon \Rightarrow a - \varepsilon \leq a_{n} \leq a + \varepsilon \end{align*} Similarly, for every $\varepsilon$, there is a natural number $N_{2}\geq 1$ such that \begin{align*} n\geq N_{2} \Rightarrow |b_{n} - b| \leq \varepsilon \Rightarrow b - \varepsilon \leq b_{n} \leq b + \varepsilon \end{align*}
Consequently, based on the given hypothesis as well as on the fact that for every $\varepsilon > 0$, there is a natural number $N = \max\{N_{1},N_{2}\}$ satisfying \begin{align*} a - \varepsilon \leq a_{n} \leq b_{n} \leq b + \varepsilon \Rightarrow b - a + 2\varepsilon > 0 \end{align*}
we can suppose that that $a > b$. Hence we can choose $\displaystyle\varepsilon = \frac{a - b}{3}$, from whence we get that \begin{align*} b - a + \frac{2(a-b)}{3} = \frac{3(b-a) + 2(a-b)}{3} = \frac{b-a}{3} < 0 \end{align*} which contradicts our assumption. Therefore the proposed result holds.
Hopefully this helps.