proof for $eleq 3(v-2)$ [Why does $d(f) geq 3$?]

Here is the well known theorem, "For planar graph $G$, if $vgeq3$ then $eleq 3(v-2)$"

I’ve reviewed my discrete mathematics note, suddenly the question crossed my mind.

When we proving the $eleq 3(v-2)$ for planar graph, $G$. We used $d(f) geq 3$.

In my note, it told me the reason why does $d(f) geq 3$ is "each face is bounded by at least three edges, but each edge borders two faces."

When I first saw it, it was clear for me. But the time passed I doubt about that.

Because… Let me suggest the counter example below.

The above graph is planar graph and $v=3, e=2,f=1$. But the $d(f) lt 3$ (I.e. the face is not bounded three edges)

Still I’ve confused what the point did I mistake.

Any help would be appreciated.