Let space $X$ be a CW complex.
We know that if $X$ contains $1$-dimensional cells (i.e the $1$-skeleton of $X$ = $X_1$ is non-void) then the $0$- skeleton $X_0$ is also non-void. This follows easily from the definition of the characteristic map for each $1$-cell.
My question now: can $X$ possess a $2$-dimensional cell (at least one) without any $1$-dimensional cells? (so that $X_1$ = $X_0$)
Thanks for your interest.
Yes, this is allowed. The only condition for the attaching map of a $2$-cell (besides, of course, continuity) is that it lands in the $1$-skeleton, and it is completely fine if this happens to be the same as the $0$-skeleton.
An example of this is one of the standard CW decompositions of the $2$-sphere $S^2$: choose the $0$-skeleton to consist of a single $0$-cell $x_0$, attach no $1$-cell, and attach a $2$-cell via the constant map to $x_0$.