A topological space $X$ is called a Fréchet–Urysohn space if for each subset $A\subseteq X$ and for each point in its closure, $x\in\overline{A}$, there is a sequence (not just a net, but a sequence) $\{a_n\}\subseteq A$ that converges to $x$: $$ a_n\underset{n\to\infty}{\longrightarrow}x. $$

Let $\varPhi$ be a set of *continuous* maps $\varphi:[0,1]\to[0,1]$ with the following property:

each sequence $\{\varphi_n\}\subseteq\varPhi$ has a subsequence $\{\varphi_{n_k}\}$ that converges to some map $f:[0,1]\to[0,1]$ pointwisely: $$ \forall t\in [0,1]\quad \varphi_{n_k}(t)\underset{k\to\infty}{\longrightarrow}f(t) $$ ($f$ is not necessarily continuous).

We consider $\varPhi$ as a set in the space $[0,1]^{[0,1]}$ of all maps $f:[0,1]\to[0,1]$ with the *topology of pointwise convergence* (in other words, we treat $[0,1]^{[0,1]}$ as the direct product of $\mathfrak{c}=\operatorname{card}([0,1])$ copies of the interval $[0,1]$).

Question:is the closure $\overline{\varPhi}$ of the set $\varPhi$ in the space $[0,1]^{[0,1]}$ a Fréchet-Urysohn space?