Hint 1: it suffices to do the case $\ell = 1$, since $D^\alpha u \in C^0$ implies that $D^{\alpha - 1}u \in C^1$.
Hint 2: Consider Hölder's inequality (or Cauchy-Schwarz if you will) for $-\infty < a < b < \infty$ $ \int_a^b |f(x)| \mathrm{d}x = \int_a^b \mathbb{1} |f(x)| \mathrm{d}x \leq \left(\int_a^b \mathbb{1}^2\mathrm{d}x \right)^{1/2}\left(\int_a^b |f(x)|^2 \mathrm{d}x\right)^{1/2} \tag{*}$ this implies that $f\in L^2(R)\implies f\in L^1_{\mathrm{loc}}(R)$ so that the function $F(x) = \int_0^x f(x) \mathrm{d}x$ is absolutely continuous by the fundamental theorem of calculus for Lebesgue integrals, and $F'(x) = f(x)$ almost everywhere. This in particular implies that $f\in H^1(R) \implies f \in C^0(R)$.
Hint 3: For the decay statement, there must be a typo: you can only prove that $\lim_{|x|\to\infty} D^\alpha u = 0$ for $\alpha < \ell$. For $\alpha = \ell$ it is false. (Let $ g(x) = \sum_{n = 1}^{\infty} n \chi_{[n,n+\frac{1}{2n^4}]}(x) $ it is evidently $L^2$ but $\lim_{|x|\to\infty} g(x)$ does not exist.)
To actually prove the statement for $\alpha < \ell$, it suffices to consider $\alpha = 0$ and $\ell = 1$ again as before. Now you can use equation (*) to show that if $f \in H^1(R)$, the total variation of $f$ in the interval $[n,n+1]$ must goes to 0 as $|n|\to\infty$. From that $f\in L^2(R)$ you then get that $\liminf_{|x|\to\infty} |f| = 0$. Put the two together you get that the limit itself must be 0.