How to prove divisibility by $7$?

I think that the comments suffice to solve this particular exercise, but more generally, any time you want to proof divisibility criteria for integers, the solution usually lies in manipulating the decimal expansion of numbers, as in (using your notation): $$\overline{abcdef}= 10^5 a+10^4 b+10^3 c +10^2 d+10^1 e + 10^0 f$$

As you can see from the comments, you can also manipulate "bigger chunks" of the expansion, as in $$\overline{abcdef}=10^4\cdot\overline{ab}+10^2\cdot\overline{cd}+10^0\cdot\overline{ef}$$

In this case, the solution comes from simply noticing that $\overline{abcdef}=10^3\cdot\overline{abc}+10^0\cdot\overline{def}$, thus giving:

$$ \overline{abcdef}=1000\overline{abc}+\overline{def}=1001\overline{abc}+(\overline{def}-\overline{abc}) $$

Since $1001$ is divisible by $7$, you get the characterization you were looking for: $\overline{abcdef}\equiv\overline{def}-\overline{abc}\mod 7$, or in other words, $\overline{abcdef}$ is divisible by $7$ if and only if $\overline{def}-\overline{abc}$ is (sign doesn't matter in this case).

Use congruences: $1000\equiv -1\mod 7$, hence the number $$[abcdef]_{10}=1000\cdot [abc]_{10}+[def]_{10}\equiv [def]_{10}-[abc]_{10}\mod7.$$

Note: This is the analog of the criterion for divisibility by $11$ in base $10$. Virtually, the numbers are written here in base $1000$.

For the exact same reason, you obtain criteria for divisibility by $13$ and by $37$.