First, let us denote $ S_n = \sum\limits_{i=1}^n\xi_i $ where $(\xi_i)_{i=1}^\infty$ is a sequence of iid, symmetrically disctributed random variables: $\mathsf E\xi_1 = 0$. Furthermore, $ X_{n+1} = X_n\cdot \mathrm e^{\mu(S_{n+1} - S_n)} $ where $S_{n+1}-S_n = \xi_{n+1}$, so $ X_{n+1} = X_n\cdot \mathrm e^{\mu\xi_{n+1}}. $ Note that $\xi_{n+1}$ is independent on $X_1,\dots,X_n$. You definition of the submartingale is correct and we are going to apply it.
We need to find $ \mathsf E[X_{n+1}|X_n,\dots,X_1] = \mathsf E\left[X_n\cdot \mathrm e^{\mu\xi_{n+1}}|X_n,\dots,X_1\right]. $
Now we apply the fact that $X_n$ is measurable w.r.t. to $\sigma$-algebra in conditional expectation and $\xi_{n+1}$ is independent of this $ \sigma$-algebra, so $ \mathsf E[X_{n+1}|X_n,\dots,X_1] = X_n\mathsf E\mathrm e^{\mu \xi_{n+1}}\geq X_n $ since $\mathrm e^x$ is a convex function and $\mathsf E\mu\xi_{n+1} = 0$, so $\mathsf E\mathrm e^{\mu \xi_{n+1}}\geq 1$ by Jensen's inequality. To complete the proof you only need to show that $\mathsf E|X_n|<\infty$.
By the way, where do we use the fact that $\mu>0$?