At first, you can show that $x_1>x_2$ easily.
If $x_{n-1}>a_n$ holds (and $0\le x_n \le 2$ for all $n$), we get
$$\begin{align}
x_{n-1}>x_n & \Longrightarrow & -x_{n-1}<-x_n \\
& \Longrightarrow & 3-x_{n-1}<3-x_n \\
& \Longrightarrow & \frac{1}{3-x_{n-1}}>\frac{1}{3-x_n} \\
& \Longrightarrow & x_n > x_{n+1}\\
\end{align}$$
so $x_n>x_{n+1}$. So this sequence convergent, by monotone convergence theorem.
Here's a more formal (and shorter) proof:
Let $(x_{n_k})$ be a convergent subsequence of $(x_n)$, which we know exists, and let $L$ be its limit. Then, given any $\varepsilon > 0$, there is some $K \in \mathbb{N}$ such that for all $k \geq K$, $|x_{n_k} - L| < \varepsilon$ (this is the definition of the limit).
Now, for any $n \geq n_K$, we will show than $|x_n - L| < \varepsilon$, which will complete the proof. For this purpose, suppose that this is not true. Then there is some $n \geq n_K$ such that $|x_n - L| \geq \varepsilon > |x_{n_K} - L|$.
First, note that if $x_1 \leq L$ then we must have $x_i \leq L$ for all $i$: if not, then we have some $i$, such that $x_1 \leq L < x_i$, but then for all $m > i$, $x_m \geq x_i > L$, so $|x_m - L| \geq |x_i - L| > 0$, so in particular, $(x_{n_k})\not\to L$, a contradiction. Symmetrically, if $x_1 \geq L$, then $x_i \geq L$ for all $i$.
Now, we have a problem: we have either $x_n \geq L + \varepsilon > x_{n_K} \geq L$ or $x_n\leq L - \varepsilon < x_{n_K} \leq L$, but then monotonicity gives us either $x_m \geq L + \varepsilon$ or $x_m \leq L - \varepsilon$ for all $m \geq n$, in particular for all but finitely many $n_k$, so $|x_{n_k} - L| \geq \varepsilon$ for all but finitely many $n_k$, so $(x_{n_k})\not\to L$, a contradiction.
Thus, we must have $|x_n - L| <\varepsilon$ for all $n \geq n_K$, hence $(x_n)\to L$.
Best Answer
Application of some sequence.
Let $x_n = b -1/n$. Then $x _n \to b^-$. Since $f$ is increasing and bounded above, so is the sequence $(f(x_n))_1^\infty$. Therefore it converges. Let the limit be $A$. By definition, for each $\varepsilon > 0$, there is some $N \in \Bbb N^*$ s.t. $\vert f(x_n) - A\vert < \varepsilon$ whenever $n \geqslant N$. Therefore for $\underline {\delta = b - x_N}$, whenever $\underline{x_N = b -\delta < x < b}$, there is an $M \in \Bbb N^*$ that $M \geqslant N$ and $x_M \leqslant x \leqslant x_{M+1}$. Then by monotonicity, $f(x_M) \leqslant f(x) \leqslant f(x_{M+1})$, and $$ \underline {- \varepsilon <}\ f(x_M) - A \leqslant \underline {f(x) - A} \leqslant f(x_{M+1}) - A\ \underline { < \varepsilon}\ , $$ i.e. $$ \lim_{ x \to b^-} f(x) = A. $$