I'll post a partial answer (it should probably be checked) to my question here, but I welcome additional answers and examples. I also sense that selective might be overkill here, and perhaps this can be done with a p-point.

My $\mathbb{N}$ is a set theorist's $\mathbb{N}$, so $\mathbb{N}=\{0,1,2,\ldots\}$ and $1=\{0\}$, $2=\{0,1\}$, etc.

Recall that an ultrafilter $\mathcal{U}$ on $\mathbb{N}$ is *selective* (or *Ramsey*) if and only if for every $2$-coloring of the pairs of $\mathbb{N}$, $\pi:[\mathbb{N}]^2\to2$, there is a homogeneous set $U\in\mathcal{U}$ for $\pi$. In addition, such ultrafilters satisfy that whenever $(U_n)$ is a decreasing sequence of sets in $\mathcal{U}$, there is a $U=\{k_0<k_1<k_2<\cdots\}\in\mathcal{U}$ (called a *diagonalization*) for which $U\setminus k_n\subseteq U_n$ for all $n$.

Let's call the following a "selective Ramsey's theorem for analysts", in the spirit of Francois Dorais' answer to Applications of infinite Ramsey's Theorem (on N)?.

**Lemma:** *Let $\mathcal{U}$ be a selective ultrafilter on $\mathbb{N}$. If $(a_{i,j})_{i,j=0}^\infty$ is an infinite matrix of real numbers such that $a_i = {\displaystyle\lim_{j\to\infty} a_{i,j}}$ exists for each $i$, and $a = {\displaystyle\lim_{i\to\infty} a_i}$ exists, then there is a set $\{k_0<k_1<k_2<\cdots\}\in\mathcal{U}$ such that $a = {\displaystyle\lim_{i<j} a_{k_i,k_j}}$.*

I'll omit the proof; I believe it's the same as the one for the non-selective version, except that you take homogenous sets in $\mathcal{U}$ at each step, and then diagonalize within $\mathcal{U}$ to get the desired set.

**Proposition**: *Let $\mathcal{U}$ be a selective ultrafilter on $\mathbb{N}$, and $(e_n)$ a Schauder basis of a Banach space $X$, with basis constant $K$. If $(x_n)$ is a normalized weakly null sequence in $X$, then there is a set $V=\{n_0<n_1<n_2<\cdots\}\in\mathcal{U}$ such that $(x_{n_k})$ is congruent to some block basic sequence $(y_k)$ of $(e_n)$. Moreover, for $\varepsilon>0$, $V$ can be chosen so that $(x_{n_k})$ has basis constant $K+\varepsilon$.*

Proof. This is based on the proof of the selection principle in Albiac & Kalton. For $i,j\in\mathbb{N}$, let $$ a_{i,j}=\max\{\|S_jx_i\|,\|x_i-S_jx_i\|\}$$ where $S_jx=\sum_{k=0}^j e_k^*(x)e_k$ for $x\in X$. Note that $\lim_{j\to\infty}a_{i,j}=0$ for all $i$. Let $U$ be as in the Lemma above, and take $0<\nu<1/4$.

Let $U_0=U\setminus N_0\in\mathcal{U}$, where $N_0$ is such that
$$\max\{\|S_jx_i\|,\|x_i-S_jx_i\|\}<\frac{\nu}{2K}$$
for all $N_0<i<j\in U$. Inductively, we define $U_{k+1}=U_k\setminus N_{k+1}\in\mathcal{U}$ so that
$$\max\{\|S_jx_i\|,\|x_i-S_jx_i\|\}<\frac{\nu^{k+2}}{2K}$$
for all $N_{k+1}<i<j\in U$.

As $\mathcal{U}$ is selective, we may choose $V=\{n_0<n_1<n_2<\cdots\}\in\mathcal{U}$ such that $U\setminus n_k\subseteq U_k$ for all $k$. Let $y_k=S_{n_{k+2}}x_{n_k}-S_{n_{k+1}}x_{n_k}$, a block sequence of $(e_n)$, thus having basis constant $\leq K$. Observe,
$$\|y_k-x_{n_k}\|=\|S_{n_{k+2}}x_{n_k}-S_{n_{k+1}}x_{n_k}-x_{n_k}\|<\frac{\nu^{k+1}}{K},$$
as $n_k<n_{k+1}<n_{k+2}$ are all in $U_k$. Hence $\|y_k\|>1-\frac{\nu}{K}\geq 1-\nu$, and
$$2K\sum_{k=0}^\infty\frac{\|y_k-x_{n_k}\|}{\|y_k\|}<2(1-\nu)^{-1}\sum_{k=0}^\infty\nu^{k+1}=2\nu(1-\nu)^{-2}<1.$$
The result follows by the principle of small perturbations. Taking $\nu$ smaller will control the basis constant of $(x_{n_k})$. QED.