$\frac{T''}{T}=\frac{R''}{R}+\frac{2}{r}\frac{R'}{R}=-\omega^2$. One can check that there are only negative eigenvalues, so I have defined it this way. Let's look at the $R$ equation first
$$rR''+2R'=(rR)''=-\omega^2 rR$$
This is a second order ODE in $rR$, with solution $rR=A\cos(\omega r)+B\sin(\omega r)$. However, for $u$ to be bounded as $r\rightarrow 0$, we must have $A=0$. This also gives some insight as to why there were only negative eigenvalues to begin with: out of all possible functions, only $\frac{\sin(r)}{r}$ does not blow up at the origin. The boundary condition implies that $\omega=n\pi$. Therefore $R=B\frac{\sin(n\pi r)}{r}$.
Then solving the $T$ equation gives $T=C\cos(n\pi t)+D\sin(n\pi t)$, but the second boundary condition forces $D$ to be $0$. The general solution is, after absorbing and redefining constants:
$$u(r,t)=\sum_{n=1}^{\infty}A_n\frac{\sin(n\pi r)}{r}\cos(n\pi t)$$
$u(r,0)=\sum_{n=1}^{\infty}A_n\frac{\sin(n\pi r)}{r}=1\Rightarrow\sum_{n=1}^{\infty}A_n\sin(n\pi r)=r\Rightarrow A_n=2\int_{0}^{1}r\sin(n\pi r)dr=\frac{(-1)^{n+1}}{n\pi}$. Finally:
$$u(r,t)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n\pi}\frac{\sin(n\pi r)}{r}\cos(n\pi t)$$
