This question concerns a step in the proof theorem 2.5. from Conway's Course in Functional Analysis. The theorem is:
If $H$ is a Hilbert space and $K$ a closed convex nonempty subset, then for any vector $h\in H$ there is a unique point $k_0$ in $K$ such that $\vert\vert h -k_0\vert\vert = \text{dist}(h,K) = \text{inf}\{\vert\vert h-k \vert\vert : k\in K \} $.
The proof starts by saying that by considering $K-h := \{k-h : k\in K \}$ instead of $K$ it suffices to assume $h=0$. I am confused why we are allowed to do this or why it allows us to assume $h=0$, since I don't see any reason why $\text{dist}(h,K-h) = \text{inf}\{\vert\vert h-k-h\vert\vert \} = \text{inf}\{\vert\vert k \vert\vert \}$ would be equal to $\text{dist}(h,K)$ in general. I'm clearly missing something.
From $\{x : x\in K-h\} = \{ k-h : k\in K \}$, it follows that $$\{\|x\| : x\in K-h \} = \{ \|k-h\| : k \in K \}$$ which implies, as stated in the comments, that $\text{dist}(0,K-h) = \text{dist}(h,K)$.
If the theorem has been proved in the case where $h = 0$, then for the general case, it can be verified that $K-h$ is a closed and convex set. So there exists a unique $x_{0}\in K-h$ such that $\|x_{0}\| = \inf \{\|x\| : x \in K-h \}$. Then there is a $k_{0}\in K$ such that $x_{0} = k_{0} - h$. Then $$\|k_{0} - h\| = \|x_{0}\| = \inf \{ \|x\| : x\in K-h \} = \inf \{ \|k-h\| : k\in K \}.$$
It can further be verified that $k_{0}$ is unique. Therefore, $k_{0}$ is the unique member of $K$ such that $\|h - k_{0}\| = \text{dist}(h,K)$.