This is probably false, but I'll ask anyway, just in case.
Conjecture 1. Suppose $E\subset[0,\infty)$ has finite Lebesgue measure. Then there is a subset $E_0$ of measure zero such that $E\setminus E_0$ admits a countable and order-respecting (in the sense that $I_k<I_{k+1}$) cover of disjoint intervals $\{I_k\}_{k=1}^\infty$, where $\bigcup_{k\in\mathbb{N}}I_k$ also has finite measure.
Since $\mathbb{R}$ is Lindelof under the Lebesgue topology, it should be enough to find an open cover $\{U_\alpha\}_{\alpha\in I}$ of $E$ such that $\bigcup_{\alpha\in I}U_\alpha$ has finite measure. Then we convert to intervals and pass to a countable subcover. Order the intervals $(I_k)_{k=1}^\infty$ according to their infimum, and throw out any $I_k$ satisfying $\sup I_k<\sup I_j$ for some $j<k$. Then relabel $I_k$ to denote $I_k\setminus\bigcup_{j<k}I_j$. And that should do the trick.
So, really, Conjecture 1 is equivalent to:
Conjecture 2. If $E\subset[0,\infty)$ has finite Lebesgue measure, then there is a subset $E_0$ of measure zero such that $E\setminus E_0$ admits an open cover $\{U_\alpha\}_{\alpha\in I}$, where $\bigcup_{\alpha\in I}U_\alpha$ also has finite measure.
To disprove Conjecture 1, it is enough to prove this anti-conjecture:
Conjecture 3. For any $\epsilon\in(0,1)$, there is a measurable subset $K_\epsilon$ of $(0,1)$ such that $\mu(K_\epsilon\cap(a,b))>0$ for any $0<a<b<1$, and $\mu(K_\epsilon)<\epsilon$. (Here, $\mu$ is the Lebesgue measure.)
If Conjecture 3 is true, then let $E=\bigcup_{j=0}^\infty(j+K_{2^{-j}})$. That will break Conjecture 1.
Best Answer
Conjecture 1 is false and in fact there is a counterexample $E$ which is an open set. Just take $E$ to be any dense open set of finite measure (to construct such a dense open set, enumerate the rationals and take an interval of length $1/2^n$ around the $n$th rational). Then $E\setminus E_0$ is still dense for any null set $E_0$. If intervals $I_k$ existed as in the conjecture, then the right endpoint of $I_k$ would have to be the same as the left endpoint of $I_{k+1}$ for each $k$ (since if they were different, $E\setminus E_0$ would fail to be dense in the interval between them), and so their union contains all but countably many points and has infinite measure.
Conjecture 2 is true, though. Indeed, for at least one definition of Lebesgue measure it is true immediately from the definition. Namely, the Lebesgue (outer) measure of a set $E$ is defined as the infimum of all sums of lengths of countable collections of open intervals that cover it. So, if $E$ has finite measure, there must exist a cover by open intervals whose union has finite measure (indeed, measure arbitrarily close to that of $E$).
Your argument deducing Conjecture 1 from Conjecture 2 is wrong; in particular, you cannot necessarily order the intervals as $(I_k)$ by their infima, since the set of the infima of the intervals may not be order-isomorphic to $\mathbb{N}$. (For instance, think about the intervals that form the complement of a Cantor set.)
Conjecture 3 is also true. Just fix an enumeration $(I_n)$ of all the intervals in $(0,1)$ with rational endpoints, and pick a subinterval $J_n\subseteq I_n$ of length at most $\epsilon/2^{n+1}$ for each $n$. Then $K_\epsilon=\bigcup J_n$ will intersect every subinterval of $(0,1)$ on a set of positive measure but $K_\epsilon$ has measure at most $\epsilon$.