I don't know of a way to use the reflection principle to prove this. It might be easier to think this way: for any walk to go from $0$ to $k$ it must, in turn, go from $0$ to $1$, then from $1$ to $2$, ..., then from $k-1$ to $k$.
Conversely, any sequence of walks from $0$ to $1$ can be joined end-to-end to form a single path from $0$ to $k$.
So,
\begin{align}
P(\cup_{n\geq 1} \{S_n\geq k\}) &= P(\text{Walk from $0$ to $1$}\;\cap\;\text{Walk from $1$ to $2$}\;\cap\;\cdots\cap\text{Walk from $k-1$ to $k$}) \\
&= P(\text{Walk from $0$ to $1$})\;P(\text{Walk from $1$ to $2$})\cdots P(\text{Walk from $k-1$ to $k$}) \\
& \qquad\qquad\qquad\qquad\qquad\qquad\text{by independence of the separate sub-walks} \\
&= P(\text{Walk from $0$ to $1$})\;P(\text{Walk from $0$ to $1$})\cdots P(\text{Walk from $0$ to $1$}) \\
&= (P[\cup_{n \geq1} \{S_n\geq 1\}])^k.
\end{align}