Let $X$ be CW-complex and by $X^n$ I denote its n-skeleton. Here is my attempt to prove that connectedness of $X$ and $X^1$ are equivalent.
First, assume that $X^1$ is not connected, which means one can represent $X^1$ as disjoint union of two closed subcomplexes. Let $X^n$ be not connected. We prove by induction that $X^{n+1}$ is not connected. Indeed, consider $X^n$ with $(n+1)$-cell attached. If one assumes that this space is connected, than he would consider the boundary of the cell attached, which is contained in $X^n$ by $C$-axiom of CW-complex. The boundary contains cells from both of disjoint components (because otherwise connectedness of the n-skeleton with cell attached would be violated), which contradicts with its connectedness. Thus, each $X^n$ is not connected and then $X$ also is.
Now let $X^1$ be connected. If one of $X^n$'s is not connected, then it can be represented as disjoint union of two subcomplexes, whose 1-skeletons (their union is $X^1$) are disjoint and closed; contradiction.