Homeomorphisms are the maps that preserve all topological properties: from a structural point of view, homeomorphic spaces might as well be identical, though they may have very different underlying sets, and if they’re metrizable, they may carry very different (but equivalent) metrics. Isometries are the analogue for metric spaces, topological spaces carrying a specific metric: they preserve all metric properties, and of course those include the topological properties. Thus, all isometries are homeomorphisms, but the converse is false.
Consider the metric spaces $\langle X,d_X\rangle$ and $\langle Y,d_Y\rangle$ defined as follows: $X=\Bbb N,Y=\Bbb Z$, $$d_X(m,n)=\begin{cases}0,&\text{if }m=n\\1,&\text{if }m\ne n\;,\end{cases}$$ for all $m,n\in X$, and $$d_Y(m,n)=\begin{cases}0,&\text{if }m=n\\1,&\text{if }m\ne n\end{cases}$$ for all $m,n\in Y$. It’s easy to check that $d_X$ and $d_Y$ are metrics on $X$ and $Y$, respectively.
Clearly these are not the same space: they have different underlying sets. However, if $f:X\to Y$ is any bijection1 whatsoever, then $f$ is an isometry between $X$ and $Y$. $\langle X,d_X\rangle$ and $\langle Y,d_Y\rangle$ are structurally identical as metric spaces: if $P$ is any property of metric spaces $-$ not just of metrizable spaces, but of metric spaces with a specific metric $-$ then either $X$ and $Y$ both have $P$, or neither of them has $P$. There is no structural property of metric spaces that distinguishes them.
What I just said about $X$ and $Y$ is true of isometric spaces in general: there is no structural property of metric spaces that distinguishes them. Considered as metric spaces, they are structurally identical, though they may have different underlying sets.
Isometric spaces may even have the same underlying set but different metrics. Consider the following two metrics on $\Bbb N=\{0,1,2,\dots\}$. For any $m,n\in\Bbb N$,
$$d_0(m,n)=\begin{cases}
0,&\text{if }m=n\\\\
\left|\frac1m-\frac1n\right|,&\text{if }0\ne m\ne n\ne 0\\\\
\frac1m,&\text{if }n=0<m\\\\
\frac1n,&\text{if }m=0<n\;,
\end{cases}$$
and
$$d_1(m,n)=\begin{cases}
0,&\text{if }m=n\\\\
\left|\frac1m-\frac1n\right|,&\text{if }m\ne n\text{ and }m,n>1\\\\
1-\frac1m,&\text{if }n=0\text{ and }m>1\\\\
1-\frac1n,&\text{if }m=0\text{ and }n>1\\\\
\frac1m,&\text{if }n=1\ne m\\\\
\frac1n,&\text{if }m=1\ne n\;.
\end{cases}$$
It’s a good exercise to show that $$f:\Bbb N\to\Bbb N:n\mapsto\begin{cases}n,&\text{if }n>1\\1,&\text{if }n=0\\0,&\text{if }n=1\end{cases}$$ is an isometry between $\langle\Bbb N,d_0\rangle$ and $\langle\Bbb N,d_1\rangle$. (HINT: Both spaces are isometric to the space $\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$ with the usual metric.) Yet these are clearly not the same space: metric $d_0$ makes $0$ a limit point of the other points, but metric $d_1$ makes $0$ an isolated point.
I don’t know of any general method for finding an isometry between isometric spaces; if you can recognize two spaces as being isometric, you probably already have a good idea of what an isometry between them must look like.
1 If you want a specific bijection, $$f(n)=\begin{cases}0,&\text{if }n=0\\\\\frac{n}2,&\text{if }n>0\text{ and }n\text{ is even}\\\\-\frac{n+1}2,&\text{if }n\text{ is odd}\end{cases}$$ does the job.
Best Answer
There is, as you say, no isometry from the plane with the uniform metric into a Euclidean space. I'm not entirely convinced the crossing diagonals are a problem, but here's an alternative proof: In the uniform metric the distance along one edge of the unit square agrees with the Euclidean distance on a number line, so the isometric image of each side of the square is a Euclidean segment. On the other hand, the uniform distance between every pair of boundary points on opposite edges of the unit square is unity, while no two Euclidean segments have this property. (Even more, there is no point in a Euclidean space lying at the same distance from every point of a Euclidean segment.)
As to your second question, the Nash embedding theorem is about Riemannian manifolds, and the plane with the uniform metric is not a Riemannian manifold, i.e., the uniform metric is not induced by a Riemannian metric.