$\newcommand\de\delta\newcommand\ep\varepsilon\newcommand\R{\mathbb R}\newcommand\N{\mathbb N}$Saúl RM [proved][2] the desired result by an application of Frostman's lemma. 

Here is an elementary proof: 

Indeed, take any $\de$ and $\ep$ in $(0,1)$. Then, by the condition $H^{n-1}(K)=0$ and the [definition of the Hausdorff measure][3], there is a set $\{Q_i\colon i\in\N\}$ of closed $n$-cubes in $\R^n$ such that $\bigcup_{i\in\N}Q_i\supseteq K$, $0<l(Q_i)<\de$ for all $i\in\N$, and 
\begin{equation*}
\sum_{i\in\N}l(Q_i)^{n-1}<\ep/2, \tag{1}\label{1}	
\end{equation*}
where $l(Q_i)$ is the edge length of the $n$-cube $Q_i$. 

For each $i\in\N$, let $N_i:=\lceil 1/l(Q_i)\rceil$, so that 
\begin{equation*}
	\frac1{l(Q_i)}\le N_i\le1+\frac1{l(Q_i)}<\frac 2{l(Q_i)}. \tag{2}\label{2}	
\end{equation*}	
For each $i\in\N$ and each $j\in[N_i]:=\{1,\dots,N_i\}$, consider the $(n+1)$-cube 
\begin{equation*}
	R_{i,j}:=Q_i\times[(j-1)l(Q_i),jl(Q_i)], 
\end{equation*}
with edge length $l(R_{i,j})=l(Q_i)<\de$. Then $\bigcup_{i\in\N}\bigcup_{j\in[N_i]}R_{i,j}\supseteq K\times[0,1]$ and 
\begin{equation*}
	\sum_{i\in\N}\sum_{j\in[N_i]}l(R_{i,j})^n
	=\sum_{i\in\N}\sum_{j\in[N_i]}l(Q_i)^n
	=\sum_{i\in\N}N_i l(Q_i)^n \\ 
	<\sum_{i\in\N}\frac 2{l(Q_i)}l(Q_i)^n
	=2\sum_{i\in\N}l(Q_i)^{n-1}<\ep, 
\end{equation*}
by \eqref{2} and \eqref{1}. 

So, again by the definition of the Hausdorff measure, 
\begin{equation}
	H^n(K\times[0,1])=0. 
\end{equation}
So/similarly, $H^n(K\times[k,k+1])=0$ for all integers $k$. 

So, by the subadditivity of the Hausdorff measure, $H^n(K\times\R)=0$. $\quad\Box$ 

(The condition that $K$ is compact was not needed or used here.)
 
[2]: https://mathoverflow.net/a/457159/36721  
[3]: https://en.wikipedia.org/wiki/Hausdorff_measure#Definition