Let denote weak convergence by $\rightharpoonup$ and norm convergence by $\to$. Let $B_E$ be the closed unit ball of $E$. Fix $\varepsilon>0$ and let $\delta$ be the modulus of uniform convexity, i.e., $\forall x,y\in B_E$,
$$
\left | \frac{x+y}{2} \right | > 1 - \delta \implies |x-y| < \varepsilon.
$$
Because $x_n \rightharpoonup x$, we get $|x| \le \liminf |x_n|$. It follows that $|x_n| \to |x|$. WLOG, we assume that $x,x_n \neq 0$ for all $n$. Let $y_n := x_n / |x_n|$ and $y := x/|x|$. Then $y,y_n \in B_E$ for all $n$. Because $x_n \rightharpoonup x$ and $|x_n| \to |x|$. Then $y_n \rightharpoonup y$ and thus $\frac{y_n+y}{2} \rightharpoonup y$. It follows that
$$
1 = |y| \le \liminf \left | \frac{y_n+y}{2} \right |.
$$
On the other hand,
$$
\limsup \left | \frac{y_n+y}{2} \right | \le \limsup \frac{|y_n| + |y|}{2}= 1.
$$
We need to prove $y_n \to y$. This is equivalent to find $N$ such that $\left | \frac{y_n+y}{2} \right | > 1 - \delta$ for all $n \ge N$. This is indeed true because $\left | \frac{y_n+y}{2} \right | \to 1$.
Fix $x\in E$ and let $(y_m)$ be a sequence that minimizes $n|x-y|^{2}+\varphi(y)$. Because $\varphi$ is proper l.s.c. convex, it is bounded from below by a continuous affine function. It follows that $(y_m)$ is bounded. On the other hand, $n|x-y|^{2}+\varphi(y)$ is convex and l.s.c. in norm topology, so it is in weak topology. Also, bounded convex set is weakly compact in reflexive space. Because $E$ is uniformly convex, it is reflexive. It follows that the minimizer indeed exists. Uniformly convex space is strictly convex, i.e., $y \mapsto |y|^2$ is strictly convex, so the minimizer is unique.
6.
We fix $x\in E$ and consider $f_n (y) : = n|x-y|^2 +\varphi(y)$. Then $f_n(y_n) \le f_n(y)$ for all $y\in E$ and $n \ge 1$. In particular, $f_{n+1}(y_{n+1}) \le f_{n+1}(y_n)$ and $f_{n}(y_{n}) \le f_{n}(y_{n+1})$, i.e.,
$$
\begin{cases}
(n+1)|x-y_{n+1}|^2+\varphi (y_{n+1}) \le (n+1)|x-y_{n}|^2+\varphi (y_{n}) \\
n|x-y_{n}|^2+\varphi (y_{n}) \le n|x-y_{n+1}|^2+\varphi (y_{n+1}).
\end{cases}
$$
This implies $\varphi (y_{n}) \le \varphi (y_{n+1})$. We have
$$
n|x-y_n|^2+\varphi(y_n) \le n|x-y|^2+\varphi(y), \quad \forall y \in E.
$$
It follows that $\varphi(y_n) < +\infty$ and thus $y_n \in D(\varphi)$ for all $n$. Also,
$$
|x-y_n|^2 - |x-y|^2 \le \frac{\varphi(y)-\varphi(y_n)}{n} \le \frac{\varphi(y)-\varphi(y_1)}{n}, \quad \forall n \ge 1, \forall y \in E.
$$
This implies
$$
\lim_n |x-y_n|^2 - |x-y|^2 \le 0, \quad \forall y \in D(\varphi).
$$
Hence
$$
\lim_n |x-y_n|^2 \le |x-y|^2, \quad \forall y \in \overline{D(\varphi)}.
$$
This means $(y_n)$ is a minimizing sequence. By 2., we get $y_n \to Px$.
Best Answer
We fix $t \in(0,1)$ and $x, y \in E$.
Let $x \neq y,|x|=|y|=1$. Then $$ |t x+(1-t) y|^2 = \varphi(tx + (1-t)y < t\varphi(x) + (1-t) \varphi(y) = t|x|^2 + (1-t)|y|^2 = 1. $$ Hence $|t x+(1-t) y| < 1$.
Let $x \neq y, x' := x/|x|$, and $y' := y/|y|$. Then $|x'|=|y'|=1$. Let $t' := t|x|/(t|x|+(1-t)|y|)$. Then $$ \left |\frac{t x+(1-t) y}{t|x|+(1-t)|y|} \right |^2 = \left | \frac{t|x| x'+(1-t) |y| y'}{t|x|+(1-t)|y|} \right |^2 = |t'x' + (1-t')y'|^2 < 1. $$
It follows that $|t x+(1-t) y|^2 < (t|x|+(1-t)|y|)^2$. Because $\mathbb R\to \mathbb R, x \mapsto x^2$ is strongly (and thus strictly) convex, we get $(t|x|+(1-t)|y|)^2 < t|x|^2 + (1-t)|y|^2$. This completes the proof.