How can I get final value of $V_{out}$ at $t=∞$ in RC circuit?

The final value theorem is for a signal, not a transfer function.

Use the transfer function to express the output signal $$ V_{mathrm{out}}(s) = frac{1}{RCs+1} V_{mathrm{in}}(s),$$ with input $V_{mathrm{in}}(s)$. Now, I assume that your input signal is a step-function $$ v_{mathrm{in}}(t) = begin{cases}0, ; mathrm{for} ; t < 0 10, ; mathrm{for} ; t geq 0end{cases},$$ with the corresponding Laplace transform $V_{mathrm{in}}(s) = mathcal{L} leftlbrace v_{mathrm{in}}(t) rightrbrace = 10frac{1}{s} $.

Now, we can apply the final value theorem $$ begin{align} lim_{sto 0} s V_{mathrm{out}}(s) &= lim_{sto 0} s frac{1}{RCs+1} V_{mathrm{in}}(s) &= lim_{sto 0} s frac{1}{RCs+1}frac{10}{s} &= lim_{sto 0} frac{10}{RCs+1}frac{s}{s} &= lim_{sto 0} frac{10}{RCs+1} &= frac{10}{RCcdot 0+1} &= 10. end{align} $$

If your input signal is an impulse (Dirac delta function) $ v_{mathrm{in}}(t) = delta(t)$ the corresponding Laplace transform $V_{mathrm{in}}(s) = mathcal{L} leftlbrace v_{mathrm{in}}(t) rightrbrace = 1 $ and the final value theorem is begin{align} lim_{sto 0} s V_{mathrm{out}}(s) &= lim_{sto 0} s frac{1}{RCs+1} V_{mathrm{in}}(s) &= lim_{sto 0} s frac{1}{RCs+1} 1 &= lim_{sto 0} frac{s}{RCs+1} &= lim_{sto 0} frac{0}{RCcdot0+1} &= 0 end{align} Which we know is correct since the system is stable because the pole $p = -frac{1}{RC} < 0$ for $R$, $C > 0$.