The topology of $C^\infty_c(U)$ with $U$ open in $\mathbb{R}^n$ is defined via the topology of $C^\infty(U)$ via a direct limit topology: you take an exhausting family of compact sets $K_i$ ($K_i \subset K_{i+1 }$ ) whose union is $U$, then take the induced topology on each of the $C^\infty(K_i)$ and take the inductive direct limit topology.
My question is: What is wrong with just taking directly the relative (induced) topology in $C^\infty_c(U)$ coming from the one in $C^\infty(U)$?
Yes, as Daniel Fischer is saying, the problem is that with that topology $C_{c}^{\infty}(U)$ would not be complete. In dimension one take $U=\mathbb{R}$, and consider a function $\phi\in C_{c}^{\infty}(\mathbb{R})$ with support in $[0,1]$ such that $\phi>0$ in $(0,1)$. Then the sequence $$ \phi_{n}(x):=\phi(x-1)+\frac{1}{2}\phi(x-2)+\cdots+\frac{1}{n}\phi(x-n),\quad x\in\mathbb{R}, $$ is a Cauchy sequence with respect to the topology induced by the one in $C^\infty(\mathbb{R})$ but its limit does not have compact support, and so it does not belong to $C_{c}^{\infty}(\mathbb{R})$.