Hint: Think about Bass's Proposition 4.14 (1): given any $\delta > 0$, there is an open set $G$ with $A \subset G$ and $m(G - A) < \delta$. Then recall (or prove) that every open set is a countable union of disjoint open intervals.
Here are some more details. An argument like yours has difficulties with the possibility $m(A) = \infty$, but we can reduce to the case $m(A) < \infty$ by intersecting $A$ with large bounded sets, as below.
Suppose $\epsilon \in (0,1)$ is such that $m(A \cap I) \le (1-\epsilon)m(I)$ for each interval $I$. Let $k$ be a positive integer and let $A_k = A \cap [-k,k]$. Note that $m(A_k) \le 2k < \infty$, and that for each interval $I$, we have $$m(A_k \cap I) \le m(A \cap I) \le (1-\epsilon) m(I). \tag{1}$$
Let $\alpha > 0$ be arbitrary, and using Proposition 4.14, choose an open set $G$ with $A_k \subset G$ and
$$m(G - A_k) < \alpha \epsilon. \tag{2}$$
In particular, $m(G) = m(A_k) + m(G-A_k) < 2k + \alpha \epsilon < \infty$.
Now $G$ can be written as a disjoint union of open intervals: $G = \bigcup_{n=1}^\infty I_n$. Note that $m(I_n) \le m(G) < \infty$ for each $n$. Also, $G - A_k = \bigcup_{n=1}^\infty (I_n - A_k)$ which is also a disjoint union.
Now since $I_n - A_k = I_n - (I_n \cap A_k)$, we have
$$m(I_n - A_k) = m(I_n) - m(I_n \cap A_k) \ge m(I_n) - (1-\epsilon)m(I_n) = \epsilon m(I_n) \tag{3}$$ using (1). So by countable additivity,
$$m(G - A_k) = \sum_{n=1}^\infty m(I_n - A_k) \ge \sum_{n=1}^\infty \epsilon m(I_n) = \epsilon m(G). \tag{4}$$
Combining (1) and (3), we get $\epsilon m(G) < \alpha \epsilon$, so $m(G) < \alpha$. In particular, since $A_k \subset G$, we have $m(A_k) < \alpha$. But $\alpha > 0$ was arbitrary, so we must have $m(A_k) = 0$.
Moreover, $k$ was arbitrary, so $m(A \cap [-k,k]) = 0$ for every $k$. Since $A = \bigcup_{k = 1}^\infty (A \cap [-k,k])$, by countable additivity we conclude $m(A) = 0$.
Now let's drop the assumption $m(A) < \infty$. For any $n$ and any interval $I$, we have $$m(A \cap [-n,n] \cap I) \le m(A \cap I) \le (1-\epsilon)m(I).$$
Hence $A \cap [-n,n]$ satisfies the same condition and moreover $m(A \cap [-n,n]) \le 2n < \infty$. So by the previous case, $m(A \cap [-n,n]) = 0$. Now since $A = \bigcup_{n=1}^\infty (A \cap [-n,n])$, by countable additivity $m(A) = 0$.
For an explicit example with a fat Cantor set, let's consider the example given on Wikipedia, where at stage $n \ge 1$ we remove $2^{n-1}$ intervals, each of length $2^{-2n}$. The final set $C$ has measure $1 - \sum_{n=1}^\infty 2^{n-1} \cdot 2^{-2n} = 1 - \sum_{n=1}^\infty 2^{-n-1} = \frac{1}{2}$.
Let $I_k$ be the leftmost interval that remains after stage $k$. At stage $k$ the leftmost interval that was removed was centered at $2^{-k}$ and had length $2^{-2k}$, so the leftmost interval that remains is
$$I_k = \left[0, 2^{-k} - \frac{1}{2} 2^{-2k}\right] = [0, 2^{-k}(1-2^{-k-1})].$$
At the next stage, we will remove from $I_k$ one interval of length $2^{-2(k+1)}$ from $I$, then two intervals of length $2^{-2(k+2)}$ and so on. So the total length of the intervals removed from $I_k$ is
$$\sum_{n=1}^\infty 2^{n-1} 2^{-2(k+n)} = 2^{-2k} \sum_{n=1}^\infty 2^{-n-1} = 2^{-2k} \frac{1}{2} = 2^{-2k-1}.$$
Therefore, we have
$$\frac{m(C \cap I_k)}{m(I_k)} = \frac{m(I_k) - 2^{-2k-1}}{m(I_k)} = \frac{2^{-k}(1-2^{-k-1}) - 2^{-2k-1}}{2^{-k}(1-2^{-k-1})} = \frac{(1-2^{-k-1}) - 2^{-k-1}}{(1-2^{-k-1})}$$
upon cancelling a factor of $2^{-k}$. It is clear by inspection that $$\lim_{k \to \infty} \frac{m(C \cap I_k)}{m(I_k)} = 1,$$ so given $\epsilon > 0$ you can choose $k$ so large that $m(C \cap I_k) > (1-\epsilon) m(I_k)$.
(Here comes my first post!)
Since the interior of E is an open set and the closure of E is a closed set, then both of them are Lebesgue Measurable (why?).
Then prove the fact that the measure of the closure of E is equal to the measure of the interior of E (you dont need this explicitly but its enlightening).
From there, and the fact that the interior is a subset of E and E is a subset of its closure, plus the definition of Lebesgue Measure for both the interior and the closure, should ring lots of bells to you.
Keep in mind the relationship between the boundary of E, its interior and its closure. Also monotony and sub-additivity of outer measure come in handy, hope it helps!
Best Answer
The inner Jordan measure of $C$ is zero, since it does not contain any intervals.
Suppose we covered $C$ by finitely many intervals $I_k$. Let $\delta$ be the minimum of the lengths of $I_k$. Choose $n$ such that during the construction of $C$, all gaps removed at the stages $n$ and later are of length less than $\delta$.
Claim: $C_n \subset \bigcup (2I_k)$, where $2I_k$ means the interval with same center and twice the length.
Proof of claim: Indeed, let $x\in C_n$. If $x\in C$, it is covered by some $I_k$. Otherwise, it is within a gap removed at some stage $\ge n$. Let $y$ be an endpoint of this gap. Note that $y\in C$ and $|x-y|<\delta$. There is $k$ such that $y\in I_k$. It follows that $x\in 2I_k$. $\quad \Box$
Since $C_n$ is the finite union of intervals, we know exactly what its Jordan measure is, and it's bounded from below by a constant independent of $n$. This gives a lower bound on $\sum |I_k|$. In conclusion, the outer Jordan measure of $C$ is positive.