Prove that this number is irrational
Consider the cases $n=10^k$. Then we get that the $n$th digit of $a$ is the $k$th digit of $\sqrt{2}$. Now, if $a$ is rational, then it repeats with some frequency, $f$. But then we can find $d$ so that $f\mid 10^{k+d}-10^k$ for $k$ large enough. Therefore, for large enough $k$, the $10^{k+d}$th digit of $a$ and the $10^k$th digit of $a$ must be the same.
But that means that $k+d$th digit of $\sqrt{2}$ is the same as the $k$th digit of $\sqrt{2}$ for large enough $k$, and therefore $\sqrt{2}$ repeats, and therefore $\sqrt{2}$ is rational, which is a contradiction.