Proof of (*)
Adding the four finite sums,
$$\sum_{k=1}^{n}H_{k}=(n+1)H_{n}-n,$$
$$\sum_{k=1}^{n}\ln{k}=\ln{n!},$$
$$\sum_{k=1}^{n}\gamma=\gamma\,n,$$
$$\sum_{k=1}^{n}\frac{1}{2k}=\frac12H_{n},$$
gives us a representation of the $n$-th partial sum for the infinite series. Writing the infinite series as the limit of partial sums, we get:
$$\begin{align}
S
&=\sum_{k=1}^{\infty}\left(H_{k}-\ln{n}-\gamma-\frac{1}{2k}\right)\\
&=\lim_{n\to\infty}\sum_{k=1}^{n}\left(H_{k}-\ln{n}-\gamma-\frac{1}{2k}\right)\\
&=\lim_{n\to\infty}\left((n+1)H_{n}-n-\ln{n!}-\gamma\,n-\frac12H_{n}\right)\\
&=\lim_{n\to\infty}\left(\left(n+\frac12\right)H_{n}-(1+\gamma)n-\ln{n!}\right).
\end{align}$$
Use Stirling's approximation for the factorial to obtain an asymptotic formula for the log-factorial term in the series:
$$n!\sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\\
\implies \ln{n!}\sim\ln{\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\right)}=\left(n+\frac12\right)\ln{n}-n+\frac12\ln{(2\pi)}.$$
Then,
$$\begin{align}
S
&=\lim_{n\to\infty}\left(\left(n+\frac12\right)H_{n}-(1+\gamma)n-\ln{n!}\right)\\
&=\lim_{n\to\infty}\left(\left(n+\frac12\right)H_{n}-(1+\gamma)n-\left(n+\frac12\right)\ln{n}+n-\frac12\ln{(2\pi)}\right)\\
&=\lim_{n\to\infty}\left(\left(n+\frac12\right)H_{n}-\gamma\,n-\left(n+\frac12\right)\ln{n}\right)-\frac12\ln{(2\pi)}\\
&=\lim_{n\to\infty}\left(n\left(H_{n}-\gamma-\ln{n}\right)+\frac12\left(H_{n}-\ln{n}\right)\right)-\frac12\ln{(2\pi)}\\
&=\lim_{n\to\infty}n\left(H_{n}-\gamma-\ln{n}\right)+\frac12\lim_{n\to\infty}\left(H_{n}-\ln{n}\right)-\frac12\ln{(2\pi)}\\
&=\lim_{n\to\infty}n\left(H_{n}-\gamma-\ln{n}\right)+\frac12\gamma-\frac12\ln{(2\pi)}\\
&=\frac12+\frac12\gamma-\frac12\ln{(2\pi)}.~~~\blacksquare
\end{align}$$
Appendix:
Using the asymptotic series for the digamma function given by Eq.16 on this Wolfram Mathworld page,
$$\begin{align}
\lim_{n\to\infty}n\left(H_{n}-\gamma-\ln{n}\right)
&=\lim_{n\to\infty}n\left(\Psi{(n+1)}-\ln{n}\right)\\
&=\lim_{n\to\infty}n\left(\frac{1}{2n}-\sum_{\ell=1}^{\infty}\frac{B_{2\ell}}{2\ell n^{2\ell}}\right)\\
&=\frac12-\lim_{n\to\infty}\sum_{\ell=1}^{\infty}\frac{B_{2\ell}}{2\ell n^{2\ell-1}}\\
&=\frac12.
\end{align}$$