Showing $\int_x^{x+1}f(t)\,dt \xrightarrow{x\to\infty}0$ for $f\in L^2 (\mathbb{R})$ 
Show that if $f\in L^2 (\mathbb{R})$ then 
  $$\lim_{x \rightarrow \infty} g(x)=0,$$
  where 
  $$g(x)=\int_x^{x+1}f(t)\,dt$$

Since $f \in  L^2 (\mathbb{R})$ then $f \in L^2[x,x+1]$ for every $x$. Then 
$$|g(x)| \leq \left(\int_x^{x+1} f(t)^2 dt\right)^{\frac{1}{2}} \int_x^{x+1} 1 dt$$
$$|g(x)|\leq \left(\int_x^{x+1} f(t)^2 dt\right)^{\frac{1}{2}}$$

Continuing from your last step, you have 
$$|g(x)|\leq\left(\int_\mathbb{R} f(t)^2\chi_{[x,x+1]}\,dt\right)^{1/2}$$
where $\chi_{[x,x+1]}(t)$ is the indicator function on the interval $[x,x+1]$.
So $$\lim_{x\to\infty}|g(x)|\leq\lim_{x\to\infty}\left(\int_\mathbb{R} f(t)^2\chi_{[x,x+1]}\,dt\right)^{1/2}$$
Since $|f(t)^2\chi_{[x,x+1]}|\leq|f(t)^2|$ which is integrable, by Lebesgue's Dominated Convergence Theorem, we can move the limit inside the integral, so 
$$\lim_{x\to\infty}\left(\int_\mathbb{R} f(t)^2\chi_{[x,x+1]}\,dt\right)^{1/2}=\left(\int_\mathbb{R} \lim_{x\to\infty}f(t)^2\chi_{[x,x+1]}\,dt\right)^{1/2}=0$$
since $\lim_{x\to\infty}\chi_{[x,x+1]}=0$
Hence $\lim_{x\to\infty}|g(x)|\leq 0$, and we can conclude from there.
