Let $X_n$ be a $F_n$ sub martingale. Suppose it exists an integrable random variable $X'$ such that $X_n \le \mathbb E(X' \mid F_n) \,\text{}a.s. \quad \forall n\in \mathbb N_0$.
First I want to show that $\sup_{n \in \mathbb N_0} \mathbb E(X_n^+) < \infty$.
$$\sup_n \mathbb E( X_n^+)\overset{\text{assumption}}{\underset{\text{}}{\le}}\sup_n \mathbb E\big(\vert\mathbb E(X' \mid F_n)\vert \big) \le \sup_n \mathbb E\big(\mathbb E(\vert X'\vert \mid F_n) \big)\overset{\text{tower}}{\underset{\text{}}{=}}\sup_n \mathbb E( \vert X' \vert)=\mathbb E(\vert X' \vert)$$ Since $ X'$ is integrable we have $\mathbb E(\vert X' \vert)< \infty$. Therefore $\sup_{n \in \mathbb N_0}E(X_n^+) < \infty $.
Furhermore I want to show that for the a.s. limit $ X_\infty \in L^1$ holds.
$$\mathbb E(\vert X_\infty \vert )=\mathbb E( \vert \lim_{n \to \infty}X_n \vert )=\mathbb E( \lim_{n \to \infty} \vert X_n \vert )\le \mathbb E(\lim_{n \to \infty}\vert \mathbb E(X' \mid F_n)\vert)=\mathbb E(\liminf_{n \to \infty}\vert\mathbb E(X' \mid F_n)\vert)\overset{\text{Fatou}}{\underset{\text{}}{\le}}\liminf_{n \to \infty}\mathbb E(\vert \mathbb E(X'\mid F_n)\vert) \le \liminf_{n\to\infty}\mathbb E(\mathbb E(\vert X' \vert \mid F_n))=\liminf_{n\to\infty}E(\vert X' \vert)=\mathbb E(\vert X' \vert)$$Again $\mathbb E(\vert X' \vert) < \infty$ by assumption.
EDIT: This only holds if $ X_n$ non negative
If $X_n$ is negative we get $$E(\vert X_\infty \vert )=E(\lim_{n\to\infty}X_n^-)=E(\liminf_{n\to\infty}X_n^-)\le \liminf_{n\to\infty} E(X_n^-)\le \liminf_{n\to\infty}E(X_n^+)- \liminf_{n\to\infty}E(X_0)=\liminf_{n\to\infty}E(\vert X_0 \vert )< \infty$$ second last line because of $\liminf E(X_n^+)=0$
Now I have trouble to show $X_n \le \mathbb E(X_\infty \mid F_n) \forall n\in \mathbb N_0$
An idea is to split $X_n$ in negative and positive parts and show that they are uniformly integrable. I am not able to continue here. Help is much appreciated and a comment about the attempt above is also welcome.
Edit2 Still one has to consider cases where $ X_n$ is negative an positive
First things first, some intuition. $X_n$ is a submartingale. This means that it tends to becoming bigger. Hence at infinity we mostly need to worry about what happens to the positive part. This can be read directly from Doob's (sub/super)martingale convergence theorem.
Now, in your first step you showed essentially that $X_n^+$ is a uniformly integrable sequence of positive random variables. It promptly follows from Doob's convergence theorem that: $$ X_n^+ \to X_{\infty}^+ \text{ a.s. and in } L_1. $$
Finally we have to deal with the less important (yet tricky) negative part. By the submartinagle property: $$ \mathbb{E}[X_n] \ge \mathbb{E}[X_0] . $$ From which we immediately deduce (since $X_n = X_n^+ - X_n^-$): $$ \mathbb{E}[X_n^-] \le \mathbb{E}[X_n^+] - \mathbb{E}[X_0] $$ Now in virtue of the first step and by applying Fatou it follows that $X_{\infty} \in L_1.$ At this point we are left with just the last step. Here we use the same treatment: we know that for $m \ge n$ $$ X_n \le \mathbb{E}[X_m | \mathcal{F}_n] = \mathbb{E}[X_m^+ | \mathcal{F}_n] -\mathbb{E}[X_m^- | \mathcal{F}_n]. $$
The first term in the sum converges in $L^1$ ans a.s. to $\mathbb{E}[X_{\infty}^+ | \mathcal{F}_n].$ For the second one we apply Fatou once again (the sign reverses the inequality).