(I think there would be better title for my question. If there is a good idea on the title, please let me know.)

Consider the following statement:

Let $A$ be a set (with the discrete topology, if necessary.) Assume that $\langle f_n\mid n<\omega\rangle$ be a sequence of function from $A$ to a Polish space $X$. If $\{f_n(a)\in X\mid n<\omega\}$ is totally bounded for all $a\in A$, then there is a subsequence $\langle f_n\mid n\in S\rangle$ (where $S$ is an infinite subset of $\omega$) such that $\langle f_n\mid n\in S\rangle$ converges.

It is known that the statement holds if $X$ is countable (Theorem 7.23 of Rudin when $X=\mathbb{C}$. The proof for general cases is similar due to total boundedness.)

Note that my statement for countable $A$ is a part of a proof of Arzela-Ascoli theorem. This statement would turn to be a kind of Arzela-Ascoli theorem for function sequences over $A$, if we require the subsequence to be uniformly convergent. However I do not know any more relation between Arzela-Ascoli theorem and my statement.

Now consider the least cardinality of $X$ that the statement fails, and let it call $\mathfrak{scp}$ (an abbreviation of subseqeunce convergenence property.) Then $\aleph_0\le \mathfrak{scp}$. In fact, we can elaborate this inequality:

Proposition. $\mathfrak{t\le scp}$, where $\mathfrak{t}$ is the tower number.

Proof. Assume that $A=\{a_\xi \mid \xi<\kappa\}$ for $\kappa<\mathfrak{t}$. We will show that every sequence $\langle f_n:A\to X\mid n<\omega\rangle$ has a convergent subsequence.

We will construct a $\subseteq^*$-decreasing sequence $\langle S_\alpha\mid \alpha < \kappa\rangle$ such that $\langle f_n(a_\xi) \mid n\in S_\alpha\rangle$ converges for all $\xi<\alpha<\kappa$.

Let $S_0=\omega$. For each $\alpha$, take any $S_{\alpha+1}\subseteq S_\alpha$ such that $\langle f_n(a_\xi)\mid n\in S_{\xi+1}\rangle$ converges. For limit $\alpha$, we can always find $S_\alpha\in [\omega]^\omega$ such that $S_\alpha\subseteq^* S_\xi$ for all $\xi<\alpha$ (as $\alpha<\mathfrak{t}$). Since $S_\alpha\setminus S_{\xi+1}$ is finite for all $\xi<\xi+1<\alpha$, we can see that $\langle f_n(a_\xi)\mid n\in S_\alpha\rangle$ converges.

Now take $S$ such that $S\subseteq^* S_\alpha$ for all $\alpha<\kappa$, then $\langle f_n(a_\xi)\mid n\in S\rangle$ converges for all $\xi<\kappa$.

Moreover, we can show that $\mathfrak{scp\le c}$: Take $A=[\omega]^\omega$ and $X=2$. Define $f_n:A\to 2$ by $$f_n(a) =\begin{cases} 0, & \text{if $n$ is the even-th number of $a$ under the increasing enumeration,} \\ 1, & \text{otherwise.} \end{cases}$$ Then we can see that $\langle f_n(a) \mid n\in a\rangle$ does not converge for all $a\in[\omega]^\omega$, so no subsequence of $\langle f_n\mid n<\omega\rangle$ converges.

My question is: is there any relation between $\mathfrak{scp}$ and any known cardinal characteristics of the continuum? For example, can we prove $\mathfrak{scp=t}$ or $\mathfrak{scp\le h}$ (where $\mathfrak{h}$ is a shattering number)? I would appreciate your help!