Let $M$ be a compact connected manifold, $X\subset M$ a closed subset, and $f:M \times [0;1] \to M$ an isotopy such that each $f\_t:M \to M$ is fixed on some open neighborhood $N\_t$ of $X$, but there are **no assumptions** on the "size" of $N\_t$ and "continuity" or "uniformity" of those neighbourhoods in $t$.



> 
> Is that true that there exists a neighborhood $N$ of $X$ such that **all** $f\_t$ are fixed on $N$?
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This question can be reformulated as follows.
For a homeomorphism $h:M \to M$ its **support** is defined as
$$
 supp(h) 
 = ( x\in M \mid h(x) \not= x )
 \equiv M \setminus Int(Fix(f)),
$$
where $Fix(h)$ is the set of fixed points of $h$, and $Int(Fix(h))$ its interior. (Please replace above the round brackets by curly ones: they are not displayed here).


Notice that the above isotopy induces a level-preserving homeomorphism $F:M\times [0;1] \to M \times [0;1]$, $F(x,t) = (f(x,t), t)$.
Then $supp(f\_t) \times \{t\} \subset supp(F)$, whence
$\mathop{\cup}\limits\_{t\in[0;1]} supp(f\_t) \times \{t\} \subset supp(F)$ as well.



> 
> The question is thus whether $$ \mathop{\cup}\limits\_{t\in[0;1]} supp(f\_t) \times \{t\} \stackrel{?}{=} supp(F).$$
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The problem is that the support of a homeomorphism is not stable under small perturbations.
Perhaps some additional assumptions on $M$ should be made.




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Also, note that $supp(f\_t) \times \{t\}$ is closed in $M\times[0;1]$.
On the other hand, the family of such sets is (a priory) not locally finite, and therefore one can not guarantee that their union is closed.
But, if $A:=\mathop{\cup}\limits\_{t\in[0;1]} supp(f\_t) \times \{t\}$ were closed, then since $A\cap (X \times [0;1]) = \varnothing$, compactness of $X \times [0;1]$ would guarantee that there exists a neighborhood $N$ of $X$ in $M$ such that $A\cap (N \times [0;1]) = \varnothing$.
Then $N$ will be the required neighborhood.



When a support of an isotopy is disjoint from a subset
Let $M=\mathbb{R}$ and let $X=\{0\}$ (edit: I see that you want the manifold to be compact, you can take $[-10,10]$ or $\frac{\mathbb{R}}{20\mathbb{\mathbb{Z}}}$ instead of $\mathbb{R}$ in the example below, in the end only what happens in a small nhood of $0$ matters).


Consider a smooth function $\varphi:\mathbb{R}^2\to\mathbb{R}$ that is $>0$ in the triangle $\{(x,y)\in\mathbb{R}^2;0<x<y<1\}$ and $0$ everywhere else. We can also suppose that $\varphi$ is $\frac{1}{2}$-Lipschitz multiplying it by a scalar.


Then I think $f\_t:\mathbb{R}\to\mathbb{R};f\_t(y)=y+\varphi(t,y)$, for $t\in[0,1]$, is a counterexample to the claim.


