Canonical projection on the set of continuous functions that equal $0$ at $x=0$ to $\mathbb R ^d $ is Lipschitz continuous? I would like to know if the following is true. 
Let $\mathcal C_{(0) } $ denote the set of continuous functions from $[0,\infty)$ to $\mathbb R^d $ that sends $0 $ in $[0,\infty)$ to $0 $ in $\mathbb R^d $ and equip $\mathcal C_{(0) } $ with the metric $$\rho(f,g):= \sum _{n=1 } ^{\infty } 2^{-n }\text{min}\{1, \sup_{0 \le t \le n } |f(t)-g(t)|\} $$ then the canonical projections $\pi_t \ :\mathcal C_{(0) } \to \mathbb R^d \ ,\pi_t(f)=f(t) $ is Lipschitz continuous (with respect to the metric $\rho $)? 

I would think it isn't since the distance between the constant zero function and the sequence of functions $f_n(t)=(mt,...,mt) $ increases without bound at the point $t $ in $\mathbb R^d $, but the book I'm reading, Brownian Motion by R. Schlling and L Partzsch, claims it is:

then later

Have I misunderstood something?
 A: I'm reading the same book. I think it's a typo, since the projection is not Lipschitz...However, it's locally lipschitz.
Here's my take on it.
We have that $|w(t)-v(t)|<\infty$, for all $t \in [0,\infty[$, since $w,v \in (\mathbb{R}^d)^I$.
$\sup_{i \in [0,n]} |w(i)-v(i)|$ is non-decreasing with $n$, and $1 \wedge \sup_{i \in [0,n]} |w(i)-v(i)|\leq 1$. So, $ \sup_{i \in [0,1]} |w(i)-v(i)|\leq \rho(w,v)\leq\sum_{i\geq 1} 2^{-i}=1$
The Locally Lipschitz definition applied to our case is:
for all $z \in C_{(0)}$, there exists $\delta_t$ such that $\displaystyle \exists_K \forall_{w,v\in B(z,\delta_t)}|\pi_t(w)-\pi_t(v)|<K\rho(w,v)$
Let's fix $t$, with $[t]$ being the smallest integer greater or equal than $t$. 
If $\sup_{i \in [0,[t]]} |w(i)-v(i)|\geq 1$, then $\displaystyle \rho(w,v)\geq \frac{1}{2^{[t]-1}}$.
So, let's choose $\displaystyle \delta_t\in \left]0,\frac{1}{2^{[t]-2}}\right[$, and we're sure that $\displaystyle \forall_{w,v\in B(z,\delta_t)}$  we have
$$\rho(w,v)\leq  \rho(w,z)+\rho(v,z)<2\cdot \frac{1}{2^{[t]-2}}=\frac{1}{2^{[t]-1}}$$
Now, we know that $|w(t)-v(t)|\leq \sup_{i \in [0,[t]]} |w(i)-v(i)|= \frac{2^{[t]}}{2^{[t]}} \sup_{i \in [0,[t]]} |w(i)-v(i)|< 2^{[t]} \rho(w,s)$
Then we just need to choose $K=2^{[t]}$
