Given $f\in L^1(\mu)$, we can have a sequence of simple measurable functions $s_n(x) \nearrow |f(x)|$ as $n\to \infty$ for every $x\in X$. Can we use Lebesgue's Monotone Convergence Theorem?
Can $|f(x)|$ be unbounded? If it can, how can we use LMCT to show that $$\int_Xs_nd\mu \to \int_Xfd\mu \text{ as } n \to \infty$$.
Yes that is absolutely allowed in any cases. (If your question is, if any $L^1$-function can be unbounded generally, consider the measure space $((0,1), \mathcal{B}((0,1)), \lambda)$ where $\mathcal{B}((0,1))$ is the Borel-$\sigma$-algebra and $\lambda$ the corresponding Lebesgue-measure. Then $f: (0,1) \rightarrow \mathbb{R}$ defined by $f(x) = \frac{1}{\sqrt{x}}$ for all $x \in (0,1)$ is unbounded and $L^1$-integrable.)
Most, if not all modern authors even use the Extended real number line $\mathbb{R} \cup \lbrace{-\infty, \infty}\rbrace$ with some special rules and restrictions, where functions my also have the value $\infty$. Even in that case the MCT is true.
No, there is something that you missed in your statement. The functions $s_n, n \in \mathbb{N}$ need to be greater or equal $0$. As a counterexample, consider $([0, \infty), \mathcal{B}([0,\infty), \lambda))$, $f \equiv 0$ and $s_n = -\chi_{[n-1, \infty)}$ for all $n \in \mathbb{N}.$