Here's an elementary way to solve 1. First the idea: The euclidean norms are quite special, since they come from inner products, and isometries in inner product spaces are very well-behaved.
Namely, we have the following result: We denote $\langle,\rangle$ inner products and $\Vert\cdot\Vert$ the respective norm.
Proposition: If $H$ and $K$ are real vector spaces with inner products, and $T:H\to K$ is an isometry, then $T$ is of the form $L=T-T(0)$ is linear and $\langle Lx,Ly\rangle=\langle x,y\rangle$ for all $x,y\in H$.
Proof: Note that $L$ is isometric and $L(0)=0$. In particular,
$$\Vert L(x)-L(y)\Vert^2=\Vert x-y\Vert^2$$
We can rewrite this as
$$\Vert Lx\Vert^2-2\langle Lx,Ly\rangle+\Vert L(y)\Vert^2=\Vert x\Vert^2-2\langle x,y\rangle+\Vert y\Vert^2$$
Now again since $L$ is isometric and $L(0)=0$, we obtain $\Vert Lx\Vert=\Vert x\Vert$, and similarly for $y$, so $\langle L(x),L(y)\rangle=\langle x,y\rangle$.
Now we calculate
\begin{align*}
\Vert L(x+y)-L(x)-L(y)\Vert^2&=\Vert L(x+y)\Vert^2+\Vert L(x)\Vert^2+\Vert L(y)\Vert^2\\
&\phantom{=} -2\langle L(x+y),L(x)\rangle-2 L(x+y),L(y)\rangle+2\langle Lx,Ly\rangle
\end{align*}
and again using the fact that $L$ preserves norm and inner product,
\begin{align*}
\Vert L(x+y)-L(x)-L(y)\Vert^2&=\Vert x+y\Vert^2+\Vert x\Vert^2+\Vert y\Vert^2\\
&\phantom{=} -2\langle x+y,x\rangle-2\langle x+y,y\rangle+2\langle x,y\rangle\\
&=0
\end{align*}
So $L(x+y)=L(x)+L(y)$. I'll leave it to you to show $L(\lambda x)=\lambda L(x)$ for $\lambda\in\mathbb{R}$ and $x\in H$. QED
Therefore, if $\mathbb{R}$ and $\mathbb{R}^2$ were isometric with their euclidean norms, there would be a linear isometry $L:\mathbb{R}^2\to\mathbb{R}$, which would imply $2=\dim\mathbb{R}^2=\dim T(\mathbb{R}^2)=\dim\mathbb{R}=1$, an absurd.
This in fact works to show that $\mathbb{R}^n$ and $\mathbb{R}^m$ are not isometric with their euclidean norms for $n\neq m$. There are several results which are a lot stronger, but they require more advanced stuff.
For $2.$ here's a hint: If $(X,d_X)$ and $(Y,d_Y)$ are metric spaces and $f:X\to Y$ is an isometry, then $d_Y(x,y)=d_X(f^{-1}(x),f^{-1}(y))$. This formula shows, in particular, that $d=d_Y$ is the only metric on $Y$ which makes $f$ an isometry (with $d_X$ on $X$).
Now if you have a bijection $f:(-1,1)^2\to\mathbb{R}^2$, use the same formula to induce a (unique) metric on $\mathbb{R}^2$ which makes $f$ an isometry.
