Let $(\Omega, \mathcal A, \mu)$ be a measurable space with a finite measure $\mu$. Let $(f_n)_{n\in\mathbb N}$ and $f$ be real measurable functions such that
$|f_n|\leq g$ for all $n\in\mathbb N$ with $g\in L^1(\mu)$
for every subsequence $(f_{n_k})_{k\in\mathbb N}$ of $(f_n)$ exists a subsequence $(f_{n'_{k}})$ of $(f_{n_k})$ such that $\lim_{k\to\infty} f_{n'_k} =f$ $\mu$-almost everywhere.
Then: $f\in L^1(\mu)$ and $\lim_{n\to\infty} \|f_n-f\|_1=0$
I think this follows directly from Lebesgue's theorem if we can show that if the above holds for subsequences of $(f_n)$, it also holds for $(f_n)$:
Suppose it doesn't hold for $(f_n)$, then we can find a $c>0$ such that $$\|f_n-f\|_1 \geq c > 0$$ for all $n\in\mathbb N$.
But according to the second property we can find a convergent subsequence $(f_{n'_k})$ such that $f_{n'_k}\to f$ $\mu$-almost everywhere. This is a contradiction and hence the proposition holds for $(f_n)$, too.
Does this reasoning hold?
You got the idea. However, notice that when you contradict the conclusion, we have the existence of $c$ such that for infinitely many $n$'s, we have $\lVert f_n-f\rVert_1\geqslant c\gt 0$.
This gives a subsequence, and then you extract a further subsequence.
It is worth mentioning that the second bullet is convergence in measure.