$\newcommand\R{\Bbb R}\newcommand\N{\Bbb N}\newcommand\conv{\operatorname{conv}}$The answer is no.

Indeed, let us use the example by Shapiro of a closed convex set in $\R^2$ such that the metric projection onto that set is not everywhere directionally differentiable.

Accordingly, for $n\in\N:=\{1,2,\dots\}$ let $a_n:=\frac\pi{2^n}$, $A:=\{a_n\colon n\in\N\}$, \begin{equation} K:=\conv(A\cup\{(0,0),(1,0)\}), \end{equation} where $\conv$ denotes the convex hull, so that $K$ is a compact convex set.

Let $x_0:=(1,0)$, so that $x_0\in\partial K$. Let $u:=(1,0)$, so that $u\cdot(x-x_0)\le0$ for all $x\in K$, where $\cdot$ denotes the dot product.

Let $z:=(0,1)$. Our $z$ and $x_0+u=(2,0)$ were respectively denoted as $d$ and $x_0$ by Shapiro.

Note that \begin{equation} z_n=\pi_{T_n}(z+nu)=n(\pi_K(x_0+u+z/n)-x_0) \end{equation} and $x_0=\pi_K(x_0+u)$. So, we can represent $z_n$ as a quotient: \begin{equation} z_n=Q(1/n),\quad Q(h):= \frac{\pi_K(x_0+u+h z)-\pi_K(x_0+u)}h. \end{equation}

Shapiro showed that there are two positive sequences $(s_j)$ and $(t_j)$ converging to $0$ such that \begin{equation} \lim_j \pi_2 Q(t_j)=:L_1\ne L_2:=\lim_j \pi_2 Q(s_j), \end{equation} where $\pi_2 a$ denotes the second coordinate of $a\in\R^2$.

Let now \begin{equation} n_j:=\lceil 1/t_j\rceil,\quad m_j:=\lceil 1/s_j\rceil, \end{equation} so that $1/n_j=t_j+O(t_j^2)$ and $1/m_j=s_j+O(s_j^2)$. Therefore and because $\pi_K$ is $1$-Lipschitz, we get \begin{equation} \lim_j \pi_2 z_{n_j}=\lim_j \pi_2 Q(1/n_j)=L_1\ne L_2=\lim_j \pi_2 Q(1/m_j)=\lim_j \pi_2 z_{m_j}. \end{equation}
Thus, the sequence $(z_n)$ does not converge. $\quad\Box$

$\newcommand\R{\Bbb R}\newcommand\N{\Bbb N}\newcommand\conv{\operatorname{conv}}$The answer is no.

Indeed, let us use the example by Shapiro of a closed convex set in $\R^2$ such that the metric projection onto that set is not everywhere directionally differentiable.

Accordingly, for $n\in\N:=\{1,2,\dots\}$ let $a_n:=\frac\pi{2^n}$, $$A:=\{(\cos a_n,\sin a_n)\colon n\in\N\},$$ \begin{equation} K:=\conv(A\cup\{(0,0),(1,0)\}), \end{equation} where $\conv$ denotes the convex hull, so that $K$ is a compact convex set.

Let $x_0:=(1,0)$, so that $x_0\in\partial K$. Let $u:=(1,0)$, so that $u\cdot(x-x_0)\le0$ for all $x\in K$, where $\cdot$ denotes the dot product.

Let $z:=(0,1)$. Our $z$ and $x_0+u=(2,0)$ were respectively denoted as $d$ and $x_0$ by Shapiro.

Note that \begin{equation} z_n=\pi_{T_n}(z+nu)=n(\pi_K(x_0+u+z/n)-x_0) \end{equation} and $x_0=\pi_K(x_0+u)$. So, we can represent $z_n$ as a quotient: \begin{equation} z_n=Q(1/n),\quad Q(h):= \frac{\pi_K(x_0+u+h z)-\pi_K(x_0+u)}h. \end{equation}

Shapiro showed that there are two positive sequences $(s_j)$ and $(t_j)$ converging to $0$ such that \begin{equation} \lim_j \pi_2 Q(t_j)=:L_1\ne L_2:=\lim_j \pi_2 Q(s_j), \end{equation} where $\pi_2 a$ denotes the second coordinate of $a\in\R^2$.

Let now \begin{equation} n_j:=\lceil 1/t_j\rceil,\quad m_j:=\lceil 1/s_j\rceil, \end{equation} so that $1/n_j=t_j+O(t_j^2)$ and $1/m_j=s_j+O(s_j^2)$. Therefore and because $\pi_K$ is $1$-Lipschitz, we get \begin{equation} \lim_j \pi_2 z_{n_j}=\lim_j \pi_2 Q(1/n_j)=L_1\ne L_2=\lim_j \pi_2 Q(1/m_j)=\lim_j \pi_2 z_{m_j}. \end{equation}
Thus, the sequence $(z_n)$ does not converge. $\quad\Box$