If $N$ is a complete linear subspace of $H$ and $\inf_{f \in N} |f-g| = d$, prove that $\exists f'$ such that $f' \in N$ and $|f'-g|=d$ The problem as stated is

Let $H$ be a Hilbert space.  If $N$ is a complete linear subspace of $H$ and $\inf_{f \in N} |f-g| = d$, prove that $\exists f'$ such that $f' \in N$ and $|f'-g|=d$.

I would like to try to prove that there exists a Cauchy sequence which converges to $f'$, but I am only able to put an upper bound of order $d$ on $|f_n-f_m|$. This is because it is possible that any two functions which are within $d+\epsilon$ of $g$ may still be up to $2d+2\epsilon$ away from each other.
I am working from the following definition of $\inf$:

If $\inf_{f \in N} |f-g| = d$ then $\forall \epsilon > 0$ $\exists$ $f'' \in N$ such that $|f''-g|<d+\epsilon$.

Edit: Ideally, the answer would prove this using a Cauchy sequence or in some other elementary way, without referring to other known results about $H$.
In other words I'm looking to show that there exists a Cauchy sequence that converges to $f'$.
 A: Let $(f_n)$ be a sequence in $N$ with $$\lim_{n\to\infty} \lvert f_n - g\rvert = d.$$
Then $(f_n)$ is a Cauchy sequence, and by continuity its limit $f'$ lies in $N$ and satisfies $\lvert f' - g\rvert = d$. And it is the unique best approximation to $g$ in $N$.
The last assertion follows by applying the argument to the sequence alternating between $f'$ and $f''$ if both $f'$ and $f''$ realise the distance.
So it remains to show that a minimising sequence is a Cauchy sequence. For that, we need that $N$ is convex (midpoint-convex suffices), and the parallelogram identity
$$\lvert x-y\rvert^2 + \lvert x+y\rvert^2 = 2\lvert x\rvert^2 + 2\lvert y\rvert^2\tag{1}$$
for all $x,y$. The identity $(1)$ follows directly from $\lvert z\rvert^2 = \langle z,z\rangle$ and the properties of an inner product.
We apply $(1)$ to $x = f_n - g$ and $y = f_m - g$ and obtain
$$\lvert f_n - f_m\rvert^2 + \lvert (f_n+f_m) - 2g\rvert^2 = 2\lvert f_n - g\rvert^2 + 2 \lvert f_m - g\rvert^2$$
and from that
$$\lvert f_n - f_m\rvert^2 = 2\lvert f_n - g\rvert^2 + 2 \lvert f_m - g\rvert^2 - 4 \left\lvert \frac{f_n+f_m}{2}-g\right\rvert^2.$$
Now, $\frac12(f_n + f_m) \in N$, so we have
$$\lvert f_n - f_m\rvert^2 \leqslant 2\lvert f_n-g\rvert^2 + 2\lvert f_m-g\rvert^2 - 4 d^2.\tag{2}$$
Since $\lvert f_n - g\rvert \to d$, for all $\varepsilon > 0$ there is an $N_\varepsilon$ with $n \geqslant N_\varepsilon \Rightarrow \lvert f_n -g \rvert^2 < d^2 + \varepsilon^2/2$. Then, for $n,m \geqslant N_\varepsilon$, $(2)$ implies $\lvert f_n - f_m\rvert < \varepsilon$, hence $(f_n)$ is a Cauchy sequence.
A: A well-known result in Hilbert spaces is that a closed convex non-empty subset of $H$ has an element of minimal norm. Your complete subspace is closed and convex and so is its translate $N - g$. Thus there exists $f'-g \in N -g$ such that $\|f'-g\|$ is minimal. 
