TL;DR: It depends.
To show that a given space $M$ is a manifold, one typically exhibits for each $x\in M$ an explicit open neighbourhood $U_x\ni x$ and an explicit homeomorphism $U_x\to \Bbb R^n$. To simplify the task, one need not specifiy different $U_x$ for each point; instead it suffices to cover $M$ with open sets (which are open nighbourhoods for each point they contain). Also, instead of a homeomorphism with all of $\Bbb R^n$ it is sufficient to specify a homeomorphism with an open subset of $\Bbb R^n$ (again, this subset contains an open ball for each of its elements and open balls are in a simple fashion homeomorphic to their containin g$\Bbb R^n$).
Example:
The set $S^1=\{\,(x,y)\in\Bbb R^2\mid x^2+y^2=1\,\}$ (with metric/topology induced from $\Bbb R^2$) is a 1-manifold.
Proof:
The set $U_N:=S^1\setminus\{(0,1)\}$ is an open subset of $S^1$. On it we can define $f_N\colon U_N\to\Bbb R$, $(x,y)\mapsto \frac{x}{1-y}$ and verify that this is indeed a homeomorphism $U_N\stackrel\approx \longrightarrow\Bbb R$. Similarly, for $U_S:=S^1\setminus\{(0,-1)\}$, we have
$f_S\colon U_S\to\Bbb R$, $(x,y)\mapsto \frac{x}{1+y}$. As $U_S\cup U_N=S^1$, we have just shown that $S^1$ is a 1-manifold. $\square$
On the other hand, showing that somehing is not a manifold may be trickier. Details depend on the specific space, of course.
Example. The set $X=\{\,(x,y)\in\Bbb R^2\mid xy=0\,\}$ is not a manifold.
Proof. The intuitive reason is the double point at $(0,0)$.
Assume $U\subseteq X$ is an open neighbourhood of $(0,0)$ and $f\colon U\to \Bbb R^n$ a homoemorphism.
Then for some $r>0$, $U$ contains the open ball $B_r((0,0))\cap X$. Under $f$, this is homeomorphic to an open subset of $\Bbb R^n$.
Now $B_r((0,0))\cap X$ has the interesting property that it is connected, but after removing the single point $(0,0)$ it becomes disconnected (into four connected components). The same should happen with $f(B_r((0,0))\cap X)$ upon removing $f((0,0))$. However, if $n>1$ then removal of a point from a connected open set produces a connected open set (because this is true for the removal of the center from an open ball). And if $n=1$ then the removal of a point from an open set produces two, not four connected components. And if $n=0$, removal of a point from a connected open set produces the empty set. Hence for all $n\in \Bbb N$ we arrive at a contradiction. We conclude that $X$ is not a manifold. $\square$
Caution. The definition for manifold that you quote may not be acceptd by everybody. As quoted, the set $S^1\cup\{(0,0)\}$ would be a manifold - which probably most would disagree with. Also, the space $\Bbb R^\times \Bbb R$ endowed with metric $$d\bigl((x,y),(x',y')\bigr)=\begin{cases}|x-x'|&\text{if }y=y'\\1+|x-x'|&\text{otherwise}\end{cases}$$ is a manifold under the quoted definition, but is not according to more generally accepted definitions. Apart from these two kinds non-manifolds that fit under your definition, there is a third kind, but as you seem to consider metric spaces (as opposed to general topological spaces) this third kind of problem does not apply.