A couple things. Since $\eta$ is in the denominator of your pde, your solution will have a problem when $\eta$ is 0. Luckily since zero is one of the end points we can pick a number close to zero without changing the answer much. A numerical solution is an approximation after all. Also your NDSolveValue
needs the limits of $\eta$ as well as $\tau$. Fixing that and choosing a very small number for zero, we have:
ClearAll["Global`*"]
β = 0.05;
zero = 10^-12;
sol = NDSolveValue[{D[U[η, τ], τ] == -Sin[τ] + β*(1/η)*D[η*D[U[η, τ], η], η],
U[η, 0] == 0, U[1, τ] == 0, Derivative[1, 0][U][zero, τ] == 0}, U, {η, zero, 1},
{τ, 0, 10}]
tp = Table[
Plot[Evaluate[sol[η, τ]], {η, zero, 1},
PlotRange -> {-2, 1}], {τ, 0, 10, .1}];
ListAnimate[tp]

In this particular case we can include $\eta = 0$ in the solution because the derivative wrt to $\eta$ is zero there. Expanding the pde we have
Derivative[0, 1][U][η, τ] == (β*Derivative[1, 0][U][η, τ])/η +
β*Derivative[2, 0][U][η, τ] - Sin[τ]
the term with $\eta$ in the denominator is $0/0$ at $\eta = 0$, so we can use L'Hospital's rule. Rewriting the pde with Piecewise functions to separate the zero case from the general case and solving we get:
Clear["Global`*"]
pde = Derivative[0, 1][U][η, τ] ==
Piecewise[{{β*Derivative[2, 0][U][η, τ], η == 0},
{(β*Derivative[1, 0][U][η, τ])/η, True}}] + β*Derivative[2, 0][U][η, τ] -
Sin[τ]
β = .05
sol = NDSolveValue[{pde, U[η, 0] == 0, U[1, τ] == 0,
Derivative[1, 0][U][0, τ] == 0},
U, {η, 0, 1}, {τ, 0, 10}];
The resulting plot looks the same as before, but we can now start the solution at $\eta = 0$.