Here is my idea. It needs a bit of refinement
Let $g$ be the decimal expansion of $\sqrt{2}$ to $m$ decimal places, where the $m$'th decimal place is not zero.
Then $g^2$ will have its final non-zero digit $d$ in the $2m$'th decimal place, i.e. $d*10^{-2m}$.
$d$ must be greater than or equal to $1$, and there are digits subsequently to d (since $\sqrt{2}$ is irrational),
so we have $g^2 < 2- 10^{-2m} $,
So $g^2+10^{-2m} < 2$
Let $(g+e)^2=2$
Then $(g+e)^2 = 2 > g^2+10^{-2m}$
So $2ge+e^2 > 10^{-2m}$
Since $e$ is very small in comparison with $ge$, we can ignore the $e^2$ term and conclude that
$e>10^{-2m}/2g = 10^{-2m}/2.828...$
This is a value which begins in the $2m+1$'th decimal place.
So consecutive zeros can run from the $m+1$'th to the $2m$'th decimal place, so we are done.