No for general topological spaces, yes for metrizable ones (and I believe the argument can be generalized to all normal spaces).
Bad example: X=a,b,c with open sets emptyset, X, a, a,b, a,c. Let A=a, then barA=X. The homotopy is given by H1=id, Htequiva for t<1. The dimension of A is 0 but the dimension of barA is 1.
On the positive side, let me begin with a quick and dirty proof in the case when barA is a compact metric space. Let Ui be an open covering of barA. We need to find a refined covering of multiplicity at most N+1 where N=dimA. It suffices to find a continuous map f:barAtoA and an open covering Vj of A such that f−1(Vj) is a refinement of Ui. Indeed, in this case we can find a refinement of Vj of multiplicity at most N+1 and its f-preimage is the desired refinement of Ui.
In the compact case, let Vj be the covering by (rho/3)-balls where rho is the Lebesgue number of the covering Ui. Then, for some t sufficiently close to 1, the map f=Ht satisfies the desired property: the preimage of every Vj has diameter less than rho and hence is contained in some of the sets Ui. Indeed, suppose the contrary. Then there is a sequence tkto1 and sequences xk,ykinbarA such that |xkyk|gerho but |Htk(xk)Htk(yk)|<2rho/3. Due to compactness we may assume that xk and yk converge to some x,yinbarA. Then |xy|gerho but |H1(x)H1(y)|le2rho/3, a contradiction.
In the general metric space case, let rho(x) denote the local Lebesgue number of Ui at x, that is the supremum of rho such that the ball Brho(x) is contained in one of the set Ui. Note that xmapstorho(x) is a positive 1-Lipschitz function on barA. It is easy to construct a continuous function u:barAto[0,1) such that the distance from x to Ht(x) is less that rho(x)/10 for all xinbarA and all t>u(x). Then the map f:barAtoA given by f(x)=Hu(x)(x) and the covering of A by the balls of the form Brho(x)/10(x), xinA, will do the job.
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