What went wrong in this proof?

What went wrong in this proof?

Let $U \subset V$ and $P_{U}v$ to be an orthogonal projection of $V$ onto
$U$. Show $\| P_Uv\| \leq \|v \|$
This geometrically makes sense.
Here is what I wrote.
Let $(e_1, \dots,e_m, e_{m+1}, \dots, e_n)$ be an orthonormal basis for
$V$. Then for any $v \in V$, we have $$v = \langle v,e_1 \rangle e_1 +
\dots+ \langle v,e_n \rangle e_n$$
and
$$\| v \|^2 = \| \langle v,e_1 \rangle \|^2 + \dots + \| \langle v,e_n
\rangle \|^2$$
Also, $P_U(v) = \langle v,e_1 \rangle e_1 + \dots + \langle v,e_m \rangle
e_m$ where we let $(e_1, \dots,e_m)$ be an orthonormal basis for $U$ and
$(e_{m+1}, \dots, e_n)$ be orthonormal base for $U^\perp$.
Then we have
$$\| v \|^2 = \| \langle v,e_1 \rangle \|^2 + \dots + \| \langle v,e_m
\rangle \|^2 + \underbrace{\| \langle v,e_{m+1} \rangle \|^2 + \dots + \|
\langle v,e_n \rangle \|^2}_{\delta^2} = \| P_U v \|^2 + \delta^2$$
Rearrange a bit $\| P_U v \|^2 = \| v \|^2 - \delta^2 \leq \| v \|^2$,
which gives the desired result