Suppose that no symbol has frequency $0$ (otherwise the claim is false).
Consider the tree $T$ built by the standard greedy algorithm to construct the Huffman code. This algorithm maintains a forest $F$ where each node $v$ is associated with a frequency $f_v$. Initially $F$ contains a collection of isolated vertices, one per input symbol (with the corresponding frequencies).
Then the algorithm greedly selects the two trees $T_1, T_2 \in F$ rooted in the vertices with minimum frequencies and replaces them with the tree obtained by merging $T_1$ and $T_2$ into a single tree via the addition of new root $r$. The frequency $f_r$ of $r$ is the sum of the frequencies of the roots of $T_1$ and $T_2$.
Let $a$ be the node corresponding to symbol $A$ and
suppose towards a contradiction that the depth of $a$ in $T$ is at least $2$.
Let $a'$ and $b$ be the parent and the sibling of $a$ in $T$, respectively.
Since $a'$ cannot be the root (it has depth $\ge 1$), it must have a sibling $x$. Some vertex $y$ of the subtree of $T$ rooted in $x$ was the root of a tree in $F$ when the isolated vertex $a$ was merged. Therefore the frequency of $y$ is at least the frequency of $a$ (othewise either $a$ or $b$ would have been merged with $y$ instead).
This is a contradiction since $f_a + f_b + f_y \ge 0.5 + f_b + 0.5 = 1 + f_b > 1$.