No, it is not. S. Franklin gave an example in his review of Wilansky's paper: take a compact space with a point $x$ that is not the limit of a non-trivial sequence, for example the ordinal $\omega_1+1$ with $x=\omega_1$, and double that point. The set of our space is $\omega_1\cup\{\omega_1^a, \omega_1^b\}$; the points other than $\omega_1^a$ and $\omega_1^b$ have their normal neighbourhoods and the (basic) neigbourhoods of $\omega_1^i$ ($i=a,b$) are $(\alpha,\omega_1)\cup\{\omega_1^i\}$ with $\alpha<\omega_1$.

The result is US and locally compact, but not Hausdorff.

No, it is not. S. Franklin gave an example in his review of Wilansky's paper: take a compact space with a point $x$ that is not the limit of a non-trivial sequence, for example the ordinal $\omega_1+1$ with $x=\omega_1$, and double that point. The set of our space is $\omega_1\cup\{\omega_1^a, \omega_1^b\}$; the points other than $\omega_1^a$ and $\omega_1^b$ have their normal neighbourhoods and the (basic) neigbourhoods of $\omega_1^i$ ($i=a,b$) are $(\alpha,\omega_1)\cup\{\omega_1^i\}$ with $\alpha<\omega_1$.

The result is US and (locally compact), but not Hausdorff.

No, it is not. Here is an example: E. K. van Douwen, An anti-Hausdorff Fréchet space in which convergent sequences have unique limits, Top. Appl. Volume 51, Issue 2, 11 June 1993, Pages 147-158

No, it is not. S. Franklin gave an example in his review of Wilansky's paper: take a compact space with a point $x$ that is not the limit of a non-trivial sequence, for example the ordinal $\omega_1+1$ with $x=\omega_1$, and double that point. The set of our space is $\omega_1\cup\{\omega_1^a, \omega_1^b\}$; the points other than $\omega_1^a$ and $\omega_1^b$ have their normal neighbourhoods and the (basic) neigbourhoods of $\omega_1^i$ ($i=a,b$) are $(\alpha,\omega_1)\cup\{\omega_1^i\}$ with $\alpha<\omega_1$.

The result is US and locally compact, but not Hausdorff.