Show that for every $T>0$, there is a smooth function with compact support $\chi : \mathbb{R}\longrightarrow[0,1]$ such that $\chi=1$ on $[-T, T]$.
OK, so I proved in a previous exercise that that if $f$ is locally integrable and $g$ is infinitely smooth with compact support, that the convolution is of class $C^{\infty}$. With that, the hint given is to consider a convolution of two functions: $f$ which is the characteristic function on some interval (which is locally integrable) and $g\in C^{\infty}_c(\mathbb{R})$. So we know the convolution would be of class $C^{\infty}$ which would take care of the smoothness aspect. I'm just not seeing how to produce $g$....
2. In the real line this problem has a nice, somewhat explicit solution: Take, for example, the standard smooth function with compact support $g$ (if there is no standard for you then just take a non-negative one with support in $[-1,1]$), and define $f(x)=\frac{\int_{-\infty}^x g(y)dy}{\int_{\mathbb{R}}g(y)dy}$, now just put $h(x)=f(x+T+1)f(T+1-x)$. Now it's standard to show that $h$ satisfies what is asked.
On the other hand, the approach you're hinted at works to construct so called "cut-off" functions over any open bounded subset, even in higher dimensions. To prove it like this, take an interval $J$ which properly contains $[-T,T]$, now let $c$ be the distance from $[-T,T]$ to the complement of $J$, $f$ the characteristic function of $J$ and take your $g$ such that its support is contained in, say, $[-c/2,c/2]$. Now it's easy to prove this functions satisfies what is asked (If you know that $\mbox{supp} f*g \subset \mbox{supp}f + \mbox{supp} g$).