Prove that: Every $\sigma$-finite measure is semifinite.

We can find $N$ such that $\mu\left(A\cap E_N\right)>0$ (otherwise, we would have for each $n$ that $\mu\left(A\cap\bigcup_{j=1}^nE_j\right)=0$ and $\mu\left(A\right)=\lim_{n\to +\infty}\mu\left(A\cap\bigcup_{j=1}^nE_j\right)$), and we have $\mu\left(A\cap E_N\right)\leqslant \mu\left( E_N\right)<+\infty$. Furthermore, $A \cap E_N\subset A$, hence the choice $F:=A\cap E_N$ does the job. This proves that $\mu$ is semi-finite.

The converse is not true: counting measure on the subsets of $[0,1]$ is semi-finite but not $\sigma$-finite.

Is it possible to think in this direction: From the definition of $\sigma-$ finite measure the $E_i$'s are finite measure since $\nu(E_i)<\infty\:\forall\,i$. Now we know from the definition that $X=\bigcup_{I=1}^{\infty}E_i$ hence for arbitrary E u pick from the sequence of $E_i, E\subset X$. We only need to show that $\nu(X)=\infty$. $\nu(X)=\nu\left(\bigcup_{I=1}^{\infty}E_i\right)$ $=\sum_{i=1}^{\infty}\nu(E_i)=\infty$.