Let $f \in L_1(X, \mu)$. Show that for every $\epsilon > 0$ there exists $\delta > 0$ such that if $E \in M$ with $\mu(E) < \delta$, then
$$\int_E |f| d\mu < \epsilon.$$
Then, find $(X,\mu)$ and a sequence $\{f_j\}\subset L_1 (X,\mu)$ for which there exists $\epsilon >0$
$$\sup_{E: \mu(E)<\delta} \sup_{j} \int_E |f_j|d\mu \geq \epsilon.$$
Proof:
Fix $\epsilon >0$, then we can find $n \in \mathbb{N}$ such that $\int_X f_n -f < \frac{\epsilon}{2}$. Choose $\delta =\frac{\epsilon}{2 n}$ Let $E \subset X$ such that $\mu (E) <\delta$. Then we can argue as follows
$$\int_E f \leq \int_E f_n + \int_E f_n-f \leq \int_E f_n + \int_X f_n-f$$
$$\int_E f \leq n \mu(E)+\frac{\epsilon}{2}=\epsilon.$$
I have no clue how to solve the other part. Any help is appreciated. Thanks!
Presumably $\delta>0$ in the last statement of the OP's is one positive number for which $$\int_E|f|\,d\mu<\varepsilon\qquad\text{if}\qquad \mu(E)<\delta$$
In such case, it suffices to consider a probability function and construct a sequence $X_n$ of random variables which is not uniformly integrable.
For example $([0,1],\mathscr{B}([0,1]),m)$ where $m$ is Lebesgue's measure. Let $f\in L_1([0,1],m)$. As the OP already checked, for any $\varepsilon>0$, there is $\delta>0$ such that \begin{align}\int_A|f|\,dm<\varepsilon\qquad\text{whenever}\qquad m(A)<\delta\tag{0}\label{zero} \end{align} Consider $f_n(t)=x_n\mathbb{1}_{[0,1/n]}(x)$ where $x_n\xrightarrow{n\rightarrow\infty}\infty$ is to be determined.
Set $\varepsilon=1$, and fix $\delta>0$ for which \eqref{zero} holds. For all $n>\delta^{-1}$,
$m\big(|f_n|>\max(1,\delta^{-1})\big)=m([0,1/n])=\frac1n<\delta$. Hence $$\sup_{\substack{E\in\mathscr{B}([0,1])\\m(E)<\delta}}\sup_n\int_E|f_n|\,dm\geq\sup_n\int_{\{|f_n|>\max(1,\delta^{-1})\}}|f_n|\,dm=\sup_n\frac{x_n}{n}$$
The sequence $x_n$ can be taken so be $x_n=2n$ for example.