Assume by contradiction that all integers are odd. Then, we have
$$2a-1 < 2^{n} \sqrt{2} < 2a$$
for some positive integer $a$. Multiplying inequalities (1) by $2$, we get $4a-2 < 2^{n+1} \sqrt{2} < 4a$. Since the integer $\left[2^{n+1} \sqrt{2}\right]$ is odd, it follows
$$4a-1 < 2^{n+1} \sqrt{2} < 4a.$$
In analogous way, after $n$ steps, we get
$$2^{n+1} a - 1 < 2^{2n} \sqrt{2} < 2^{n+1} a$$
From the left inequality above it follows
$$a - 2^{n-1} \sqrt{2} < \frac{1}{2^{n+1}}$$
hence
$$\frac{a^{2} - 2^{2n-1}}{a + 2^{n-1} \sqrt{2}} < \frac{1}{2^{n+1}}$$
From the right inequality in the first step we have $2a > 2^{n} \sqrt{2}$ hence $a^{2} > 2^{2n-1}$, that is $a^{2} - 2^{2n-1} \geq 1$. Using this inequality, from above it follows
$$\frac{1}{a + 2^{n-1} \sqrt{2}} < \frac{1}{2^{n+1}}$$
hence
$$2^{n+1} < 2^{n-1} \sqrt{2} + a < 2^{n-1} \sqrt{2} + \frac{1}{2} (2^{n} \sqrt{2} + 1)$$
From this we get $2^{n+2} < 2^{n+1} \sqrt{2} + 1$, that is
$$2 < \sqrt{2} + \frac{1}{2^{n+1}} < \frac{3}{2} + \frac{1}{2} = 2$$
contradiction.