$\mathbf {The \ Problem \ is}:$ Let, $X$ be a simply connected based CW space with $H_k(X)=0$ for all $k>c.$ Show $X$ is homotopy equivalent to a CW space $\Gamma X$ with no $t$-cells for $t>(c+1).$
$\mathbf {My \ approach}:$ Let $X^c$ be the $c$-th skeleton of $X.$ As $\pi_1(X)=0$ and the inclusion $X^c \hookrightarrow X$ induces isomorphism on $\pi_j$ for all $j<c$ then $X^c$ is also simply connected .
Now, $(X,X^c)$ is a based CW pair which is $c$-connected and $\pi_1(X^c)=0$ then by relative Hurewicz thm., $\pi_{c+1}(X,X^c)\cong H_{c+1}(X,X^c).$ Now, from LES of homology groups of pair $(X,X^c)$ we get a SES :
$0\longrightarrow H_{c+1}(X,X^c)\longrightarrow H_c(X^c)\longrightarrow H_c(X)\longrightarrow 0$
but I can't proceed anywhere from here.
If we can show there is a $f_*$ inducing isomorphism only all homology groups of a CW space $\Gamma X$ with that of $X,$ then $f$ will be homotopy equivalence .
Any hints ? Thanks in advance.
I guess a relatively high-tech approach that immediately implies the result would be to use the "Moore decomposition" which is Eckmann-Hilton dual to the Postnikov decomposition. It gives a sequence of cofibrations $\{X_n\hookrightarrow X_{n+1}\}_{n\ge 2}$ between CW complexes with the property:
$$ X_n\hookrightarrow X_{n+1}\to M(H_{n+1}(X),n+1),\quad X_2=M(H_2(X),2). $$
is a cofiber sequence. Here $M(H_{n+1}(X),n+1)$ is the Moore space.
One then recovers $X$ (up to homotopy equivalence of course) as $\operatorname{colim} X_n.$ This is spelled out in $\S$ 2.7 of Baues's "Homotopy Type and homology". There it is shown that such sequence can be constructed by consecutively attaching $M(G,n)$'s along homotopy trivial maps. In particular, this implies that the space $X_c$ satisfies your requirement (i.e. it has no cells above dimension $c+1$ and is homotopy equivalent to $X$ by Whithead's theorem).
However, a concrete construction (probably the one alluded to in the comments) that follows from the method described above is also possible (it is in fact verbatim the proof of Baues's Theorem 2.7.4). Consider a short exact sequence:
$$ 0\to Z_{c+1}\overset{i}{\to} C_{c+1}\overset{d_c}{\to} B_c\to 0 $$
This sequence splits since $B_c$ is projective and we can choose a section $t:B_c\to C_{c+1}$ of the map $d_c.$ Then we have a commutative diagram:
It remains to observe that by taking $\operatorname{Cone}(f_B)=X_c$ we get a space that is homotopy equivalent to $X$ and has no cells above dimension $c+1.$ The latter follows from the standard construction of the Moore complex (e.g. here), observe that over $\mathbb{Z}$ the group $B_c$ is free and hence $M(B_c,c)$ has only cells in dimension $\le c$.