I need to show that if $c\in R\!\left(X^T\right),$ then $c\in R\!\left(X^T X\right).$ Here, assume that $X$ is a finite matrix, not necessarily square, and that $c$ is a vector of the appropriate shape to make the matrix multiplications valid. Also, for notation: $R\!\left(X^T\right)$ is the range space of $X^T,$ which in turn is the transpose matrix of $X.$ Assume everything in sight is real, not complex.
I know that $c\in R\!\left(X^T\right)$ if and only if there exists $z$ s.t. $X^Tz=c.$ Also, $c\in R\!\left(X^T X\right)$ if and only if there exists a vector $\lambda$ such that $c=X^TX\lambda.$ So it appears, somehow, that I must show $z=X\lambda.$ It doesn't seem obvious to me that $z$ should be in the range of $X,$ which is essentially what that's saying.
Important note: I just realized that if $X$ is not square, then we have the possible objection that the dimensions don't work out. That is, the number of components in a vector in $R\!\left(X^T\right)$ is not necessarily the same as the number of components in a vector in $R\!\left(X^TX\right).$ (Notice I'm not using the term "dimension" here - that's deliberate.) That's certainly true, but let's assume that this is not an objection - which, for all I know, might be tantamount to saying that $X$ is square. Or maybe we can simply consider that $X\lambda$ can be considered equal to $z$ if we can lop off enough components from one or the other such that what remains is equal.
How do I continue?
This question has come up in various forms many times on this site. In general, we know that $R(AB) \subset R(A)$ for any matrices $A$ and $B$. However, in your case, you get the reverse inclusion by dimension considerations.
The claim is that $R(X^\top) = R(X^\top X)$ because you have the inclusion $R(X^\top X)\subset R(X^\top)$ and the two subspaces have the same dimension. To see this, we apply the nullity-rank theorem. We observe instead that $N(X)\subset N(X^\top X)$ (why?) and note that if $X^\top Xv = 0$, then $0 = X^\top Xv\cdot v = Xv\cdot Xv = \|Xv\|^2$, so $Xv=0$. This shows that $N(X^\top X)\subset N(X)$ and therefore that $N(X^\top X) = N(X)$. It follows from nullity-rank that $\text{rank} (X^\top X) = \text{rank}(X)$. Since $\text{rank} (X^\top) = \text{rank}(X)$, we have $ \text{rank}(X^\top) = \text{rank}(X^\top X)$, so $\dim(R(X^\top)) = \dim(R(X^\top X))$, as required.