Prove that $\|u\|_{L^\infty(\mathbb R)} \le C\|u\|_{H^1(\mathbb R)}$ 
Use the Fourier transform to prove that if $u \in H^s(\mathbb R^n)$ for $s > \frac n2$, then $u \in L^\infty (\mathbb R^n)$, with the bound $$\|u\|_{L^\infty(\mathbb R^n)} \le C\|u\|_{H^s(\mathbb R^n)}$$ for a constant $C$ depending only on $s$ and $n$.

(from PDE Evans (2nd edition), Chapter 5, Exercise 21)
First, I want to reduce this problem to its analogue in $\mathbb R$ only ($n=1$); it will be easier for me to solve this problem in this way. I will get back to the original problem afterwards.

Use the Fourier transform to prove that if $u \in H^1(\mathbb R)$, then $u \in L^\infty (\mathbb R)$, with the bound $$\|u\|_{L^\infty(\mathbb R)} \le C\|u\|_{H^1(\mathbb R)}$$ for a constant $C$.

First, I note that $H^1(\mathbb R)=W^{1,2}(\mathbb R)$, the Sobolev space of the functions $u,u' \in L^2(\mathbb R)$
Question: 
What an I do to estimate $\|u\|_{L^\infty(\mathbb R)}$ using $\|u\|_{H^1(\mathbb R)}$ as part of the bound? (Basically, I need to prove that $\|u\|_{L^\infty(\mathbb R)} \le C\|u\|_{H^1(\mathbb R)}$.)
Necessary context:
Here is the definition from the Fourier transform (from page 187 of the textbook) adjusted from $\mathbb R^n$ to $\mathbb R$:

DEFINITION. If $u \in L^1(\mathbb R)$, we define its Fourier transform $\mathcal Fu=\hat u$ by $$\hat u(y) := \frac 1{(2\pi)^{1/2}} \int_{\mathbb R} e^{-ix \cdot y} u(x) \, dx$$ and its inverse Fourier transform $\mathcal F^{-1}u=\bar{u}$ by $$\bar u(y) := \frac 1{(2\pi)^{1/2}} \int_{\mathbb R} e^{ix \cdot y} u(x) \, dx.$$ Since $|e^{\pm ix \cdot y}|=1$ and $u \in L^1(\mathbb R)$, these integrals converge for each $y \in \mathbb R$.

(Note that in the textbook an upside-down hat over $u$ is used for the inverse Fourier transform. Here instead I used a bar over $u$ because I can't find the MathJax code for an upside-down hat.)
Extending these definitions to $u \in L^2(\mathbb R)$, we have

THEOREM 1 (Plancherel's Theorem). Assume $u \in L^1(\mathbb R) \cap L^2(\mathbb R)$. Then $\hat u, \bar u \in L^2(\mathbb R)$ and $$\|\hat u\|_{L^2(\mathbb R)}=\|\bar u\|_{L^2(\mathbb R)}=\|u\|_{L^2(\mathbb R)}.$$

 A: Step 1: Assume $u \in W^{1,1}(\mathbb{R}) \cap H^1(\mathbb{R}) \cap C^\infty(\mathbb{R})$, then $$u(x) = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{ix\xi}\hat{u}(\xi)\, d\xi,$$ so that by Holder's inequality, Minkowski's inequality and Plancherel's identity we have
\begin{align}
|u(x)| \le &\ \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}|\hat{u}(\xi)|\, d\xi \\
= &\ \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\frac{\color{blue}{\sqrt{1 + |\xi|^2}}}{\sqrt{1 + |\xi|^2}}\color{blue}{|\hat{u}(\xi)|}\, d\xi \\
\le &\ \frac{1}{\sqrt{2\pi}}\Big(\color{green}{\int_{\mathbb{R}}\frac{1}{1 + |\xi|^2}\, d\xi}\Big)^{\frac12}\Big(\int_{\mathbb{R}}(1 + |\xi|^2)|\hat{u}(\xi)|^2\, d\xi\Big)^{\frac12} \\
\le &\ C\Big(\int_{\mathbb{R}}(1 + |\xi|^2)|\hat{u}(\xi)|^2\, d\xi\Big)^{\frac12} \\
\le &\ C\big(\|\hat{u}\|_{L^2(\mathbb{R})} + \||\xi|\hat{u}\|_{L^2(\mathbb{\mathbb{R}})}\big) \\
= &\ C\big(\|u\|_{L^2(\mathbb{R})} + \|u'\|_{L^2{(\mathbb{R})}}\big).
\end{align}
Take the supremum over all $x$ to obtain the desired estimate. The step colored in $\color{blue}{\text{blue}}$ is the one you need to modify in order to extend the result to higher dimensions: you need the integral in $\color{green}{\text{green}}$ to be finite.
Step 2: To remove the $C^{\infty}$- assumption apply the Meyer-Serrin approximation theorem. Finally use that $L^1(\mathbb{R}) \cap L^2(\mathbb{R})$ is dense in $L^2(\mathbb{R})$ to remove the assumption that $u \in W^{1,1}(\mathbb{R}).$ Since you are reading chapter $5$ in Evan's book you should be able take it from here.
