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I'm trying to establish the existance of $u in K subset V$, a closed convex subset of vector space with an inner product, such that for a fixed $q in V$:
$$(u-q,v-u) geq 0 quad forall v in K$$
The proof I'm reading proceeds as follows:

Note that the inequality holds iff $|u-q| leq |v-q|$. Since $K$ is a closed convex subset, there exists a unique element of $K$ that minimizes $|v-q|$, namely $u = P_Kq$ where $P_K$ is the projection of $q$ onto $K$.

This was a bit terse for me. Why does the inequality hold? From what fact does $u = P_Kq$ follow?

There's an error in the first statement of the proof. The variational inequality proves the inequality in norms, but the converse simply doesn't hold. For example, in $BbbR^2$ under the dot product, take $u = (0, 0)$, $v = (0, 3)$, and $q = (1, 1)$. Then
$$|u - q|^2 = 2 le 5 = |v - q|^2,$$
but
$$langle u - q, v - u rangle = (-1, -1) cdot (0, 3) = -3 < 0.$$
I suggest finding another proof of this inequality.

But, as for your second question, if you can show that $|u - q| le |v - q|$ for all $v in K$ (and where $u$ is assumed to be in $K$), then you've found a point $u in K$ that is of minimal distance from $q$. This, by definition, makes $u$ the metric projection of $q$ onto $K$.

EDIT: Actually, I have a proof of this inequality that I wrote up on hand:

Theorem: Suppose $X$ is a real Hilbert space, and $C$ is closed, non-empty, and convex. Let $x in X$ and $z in C$. Then,
$$langle x - z, c - z rangle le 0 quad forall , c in C$$
if and only if $z = p_C(x)$.

Proof: Suppose $z, c in C$, with $z neq c$. Let $f : mathbbR to mathbbR$ be defined by
beginalign*
f(lambda) &= |x - lambda c - (1 - lambda)z|^2 - |x - z|^2 \
&= 2 lambda langle x - z, z - c rangle + lambda^2 |z - c|^2.
endalign*
Note that this is a convex quadratic in $lambda$.

Suppose that $z = p_C(x)$. When $lambda in [0, 1]$, we have $lambda c + (1 - lambda)z in C$, hence $f(lambda) ge 0 = f(0)$. The minimum value of $f$ must be achieved at some $lambda^* le 0$. Thus,
$$0 ge lambda^* = frac-langle x - z, z - c ranglez - c iff langle x - z, c - z rangle le 0.$$
Note that the final inequality also holds for when $c = z = p_C(x)$.

Conversely, suppose $z in C$ such that $langle x - z, c - z rangle le 0$ for all $c in C$. Therefore, when $c neq z$, $lambda^* le 0$. This implies that $f$ is increasing on the interval $[0, 1]$. Hence,
$$|x - c| = f(1) ge f(0) = |x - z|.$$
As this holds for arbitrary $c in C$, we have $z = p_C(x)$.

The statement is incorrect. Consider $V = mathbb R^2$ and $K = overlineB_1(0)$, $u=(0,0)$, $v=(epsilon,1-epsilon)$ for some small $epsilon > 0$ and $q = (2,0)$. Then $|u-q|le|v-q|$ and $(u-q,v-u) = -2epsilon < 0$.

I think the proof is missing an important detail: What we actually have is that for $u in K$ and $q in V$ the statement
$$ (u - q, v - u ) ge 0 qquad forall v in K$$
is equivalent to
$$ |u - q | le | v - q | qquad forall v in K.$$

The proof is given in the answer by Theo Bendit.

The second statement is just $u = operatornameproj_K(q)$.