[proofplan]
Write $p_i=\mu(A_i)$, so $(p_1,\dots,p_k)$ is a probability vector with strictly positive entries. Non-negativity follows term by term from $0<p_i\le 1$. The upper bound follows from the elementary logarithmic inequality $\log t\le t-1$, applied to $t=1/(kp_i)$ and summed with weights $p_i$. The equality cases are exactly the equality cases in these two estimates.
[/proofplan]
[step:Convert the partition into a positive probability vector]
For each $i\in\{1,\dots,k\}$, define
\begin{align*}
p_i:=\mu(A_i).
\end{align*}
Because $\alpha$ is a measurable partition of the probability space $(X,\mathcal B,\mu)$, the sets $A_1,\dots,A_k$ are pairwise disjoint, belong to $\mathcal B$, and satisfy $\bigcup_{i=1}^k A_i=X$. Hence finite additivity gives
\begin{align*}
\sum_{i=1}^k p_i=\sum_{i=1}^k \mu(A_i)=\mu(X)=1.
\end{align*}
The hypothesis that the atoms are non-null gives $p_i>0$ for every $i$, and the identity above gives $p_i\le 1$ for every $i$. Therefore
\begin{align*}
H_\mu(\alpha)=-\sum_{i=1}^k p_i\log p_i.
\end{align*}
[/step]
[step:Prove non-negativity and characterize equality]
Since $0<p_i\le 1$, we have $\log p_i\le 0$ for every $i\in\{1,\dots,k\}$. Therefore each summand satisfies
\begin{align*}
-p_i\log p_i\ge 0.
\end{align*}
Summing over $i$ gives
\begin{align*}
H_\mu(\alpha)=-\sum_{i=1}^k p_i\log p_i\ge 0.
\end{align*}
Equality holds if and only if $-p_i\log p_i=0$ for every $i$. Since $p_i>0$, this is equivalent to $\log p_i=0$ for every $i$ with positive mass, hence to $p_i=1$ for such an atom. Because $\sum_{i=1}^k p_i=1$, this occurs exactly when one atom has measure $1$ and all other atoms have measure $0$; under the present non-null indexing, this is possible only when $k=1$.
[/step]
[step:Bound the entropy above by $\log k$ using the logarithmic inequality]
We use the elementary inequality
\begin{align*}
\log t\le t-1
\end{align*}
for every $t>0$, with equality if and only if $t=1$. For each $i\in\{1,\dots,k\}$, apply this inequality to
\begin{align*}
t_i:=\frac{1}{kp_i}>0.
\end{align*}
Then
\begin{align*}
-\log(kp_i)=\log\left(\frac{1}{kp_i}\right)\le \frac{1}{kp_i}-1.
\end{align*}
Multiplying by $p_i>0$ gives
\begin{align*}
-p_i\log(kp_i)\le \frac{1}{k}-p_i.
\end{align*}
Summing over $i$ yields
\begin{align*}
-\sum_{i=1}^k p_i\log(kp_i)\le \sum_{i=1}^k\left(\frac{1}{k}-p_i\right)=1-1=0.
\end{align*}
Since
\begin{align*}
-\sum_{i=1}^k p_i\log(kp_i)= -\sum_{i=1}^k p_i\log p_i-\sum_{i=1}^k p_i\log k=H_\mu(\alpha)-\log k,
\end{align*}
we obtain
\begin{align*}
H_\mu(\alpha)\le \log k.
\end{align*}
[guided]
The goal is to compare the entropy with $\log k$, the entropy of the uniform probability vector. To make this comparison algebraic, rewrite the difference $H_\mu(\alpha)-\log k$ as a sum involving $kp_i$.
We use the elementary logarithmic inequality
\begin{align*}
\log t\le t-1
\end{align*}
for every $t>0$, with equality if and only if $t=1$. For each index $i\in\{1,\dots,k\}$, define the positive number
\begin{align*}
t_i:=\frac{1}{kp_i}.
\end{align*}
This is valid because every atom is non-null, so $p_i>0$. Applying the inequality to $t_i$ gives
\begin{align*}
\log\left(\frac{1}{kp_i}\right)\le \frac{1}{kp_i}-1.
\end{align*}
Since $\log(1/(kp_i))=-\log(kp_i)$, we get
\begin{align*}
-\log(kp_i)\le \frac{1}{kp_i}-1.
\end{align*}
Multiplying by the positive number $p_i$ preserves the inequality:
\begin{align*}
-p_i\log(kp_i)\le \frac{1}{k}-p_i.
\end{align*}
Now sum this estimate over all atoms. The right-hand side collapses because the $p_i$ form a probability vector:
\begin{align*}
-\sum_{i=1}^k p_i\log(kp_i)\le \sum_{i=1}^k\left(\frac{1}{k}-p_i\right)=1-\sum_{i=1}^k p_i=0.
\end{align*}
Finally expand the logarithm:
\begin{align*}
-\sum_{i=1}^k p_i\log(kp_i)= -\sum_{i=1}^k p_i\log p_i-\sum_{i=1}^k p_i\log k=H_\mu(\alpha)-\log k.
\end{align*}
Thus
\begin{align*}
H_\mu(\alpha)-\log k\le 0,
\end{align*}
which is the desired upper bound
\begin{align*}
H_\mu(\alpha)\le \log k.
\end{align*}
[/guided]
[/step]
[step:Identify the equality case for the upper bound]
The upper bound was obtained by summing the inequalities
\begin{align*}
-p_i\log(kp_i)\le \frac{1}{k}-p_i.
\end{align*}
Since each inequality comes from $\log t\le t-1$, equality in the sum holds if and only if equality holds for every $i\in\{1,\dots,k\}$. The equality condition for the logarithmic inequality is $t_i=1$, so equality holds if and only if
\begin{align*}
\frac{1}{kp_i}=1
\end{align*}
for every $i\in\{1,\dots,k\}$. This is equivalent to
\begin{align*}
p_i=\frac{1}{k}
\end{align*}
for every $i$. Recalling $p_i=\mu(A_i)$, we conclude that $H_\mu(\alpha)=\log k$ if and only if every atom of $\alpha$ has measure $1/k$.
[/step]
