Uniform integrability of RV's $\mathbf{Theorem}$: Let $Y \in \mathbb{L}_1$, then the RV $(\mathbb{E}[Y \mid \mathcal{F}], \mathcal{F} \subset \mathcal{A} \space \sigma\text{-algebra})$ are uniformly integrable. 
$\mathbf{Proof}$: Choose $K \in (0,\infty)$:
\begin{align}
\mathbb{E}[\mathbb{E}[Y \mid \mathcal{F}] \mathbf{1}_{|\mathbb{E}[Y \mid \mathcal{F}]| \ge K}] &\le \mathbb{E}[\mathbb{E}[Y \mid \mathcal{F}] \mathbf{1}_{\mathbb{E}[|Y| \mid \mathcal{F}] \ge K}] \\
&=\mathbb{E}[Y  \mathbf{1}_{\mathbb{E}[|Y| \mid \mathcal{F}] \ge K}]
\end{align}
Next, we know, that 
\begin{align}
\mathbb{P}(\mathbb{E}[Y \mid \mathcal{F}] \ge K) &=\int_\Omega \mathbf{1}_{\mathbb{E}[Y \mid \mathcal{F}] \ge K}] \, d \mathbb{P}\\
&\le \frac{\mathbb{E}[\mathbb{E}[Y \mid \mathcal{F}]]}{K} = \frac{\mathbb{E}[|Y|]}{K}
\end{align}
($\mathbf{Q}$: How do we know that? Is it an application of the Martingale inequality - generalized Kolmogorov for $\psi(K) =K$ and $U = Y$ ?) My guess is no - then by what property do we obtain the above inequality?)
Hence,
$$\sup\left\{ \mathbb{E}[\mathbb{E}[Y \mid \mathcal{F}] \mathbf{1}_{|\mathbb{E}[Y \mid \mathcal{F}]| \ge K}] \right\} \le \sup \left\{\mathbb{E}[|Y| \mathbf{1}_A ]: \mathbb{P}(A) \le \frac{\mathbb{E}[|Y|]}{K}\right\}$$
And uniform integrability follows from uniform continuity of expectation.

I think you're missing some absolute values throughout your post. The inequality is just Markov's inequality applied to the non-negative variable ${\rm E}[|Y|\mid\mathcal{F}]$. For completeness: If $X$ is any non-negative random variable and $K>0$, then $$X\geq K \iff \frac{X}{K}\geq 1$$ and hence
$$
1_{\{X\geq K\}}\leq \frac{X}{K}1_{\{X\geq K\}}\leq \frac{X}{K}.
$$
Apply this to $X={\rm E}[|Y|\mid\mathcal{F}]$ and integrate both sides.
