Note. This is an easy consequence of the Sylow Theorems as well: given any finite group $G$, a prime $p$, and if $p^n$ is the largest power of $p$ that divides $|G|$, then given any subgroup $H$ of $G$ such that $|H|=p^i$ with $i\lt n$, there exists a subgroup $K$ of $G$ that contains $H$ and has order $p^{i+1}$. That $H$ is normal in $K$ will follow because the index of $H$ in $K$ is the smallest prime that divides $K$. Applying this to the case of $G$ of order $p^n$ gives you the desired conclusion.
But if you are trying to prove this in order to prove the Sylow Theorems (or you don't know this version of the Sylow Theorems), you can prove it directly, for example, as follows:
Let $|G|=p^m$. We proceed by induction on $m$.
If $m=1$, then either $H$ is trivial and we can take $H_0=H$, $H_1=G$; or else $H=G$ and we can take $H_0=G$.
Assume the result is true for any group of order strictly less than $m$. Let $H$ be a subgroup of $G$.
If $Z(G)\cap H$ is nontrivial, then consider $H/(H\cap Z(G))$ in $G/(H\cap Z(G))$. By induction, we can find a subnormal chain $\frac{H}{H\cap Z(G)} = H_0 \triangleleft H_1\triangleleft\cdots\triangleleft H_n = \frac{G}{H\cap Z(G)}$ with $|H_{i+1}/H_{i}|=p$. By the isomorphism theorems, there exist subgroups $\widehat{H_i}$ of $G$, with $\widehat{H_i}/(H\cap Z(G)) = H_i$, and we have $\widehat{H_i}\triangleleft \widehat{H_{i+1}}$, $[\widehat{H_{i+1}}:\widehat{H_i}] = [H_{i+1}:H_i]=p$, giving the desired chain $H = \widehat{H_0}\triangleleft \widehat{H_1} \triangleleft\cdots\triangleleft \widehat{H_n}=G.$
If $Z(G)\cap H=\{1\}$, then let $g\in Z(G)$ be an element of order $p$ (always possible, since $Z(G)\neq\{1\}$). Let $H'=\langle H,g\rangle$. Then $H\triangleleft H'$, and $|H'/H| = p$. Now we have $H'\cap Z(G)\neq\{1\}$, so we can apply the previous case to $H'$ to obtain a chain: $H'=H_0\triangleleft H_1\triangleleft\cdots\triangleleft H_n = G,$ with $|H_{i+1}/H_i|=p$. Now we have: $H = H_{-1}\triangleleft H_0\triangleleft\cdots\triangleleft H_n=G$ as a chain with the properties we want.