Set of functions on $C[0,1]$ with bounded derivative is compact

Question

Just want to check if my proof works or not.

Question: Given $K > 0$, determine if the set \begin{equation*} Y = \{f \in C^1[0,1], f(0) = 0, \Vert f'\Vert_\infty\leq K\} \end{equation*} is compact in $(C[0,1],\Vert \cdot \Vert_\infty)$.

My attempt: if $f' \leq K$ for all $ x \in (0,1)$, then $Y$ is equicontinuous (by using $\delta(\epsilon) = \epsilon/K$ for each $\epsilon > 0$).
Also, since for every $f \in Y$, $\Vert f \Vert_\infty \leq \Vert K(\cdot) \Vert = K$, then $Y$ is also bounded.
Any convergent sequence $(f_n)$ in $Y$ converges in $Y$ (I think), making $Y$ closed.
Hence by Ascoli's theorem, $Y$ is compact.

Answer 1

(Rewriting @SeverinSchraven's comment as an answer)

Take the function $g: \mathbb{R} \to \mathbb{R}$, defined by
\begin{equation*} g(x) = \begin{cases} 0, \quad x < 0,\\ K(2x^2-x^3), \quad 0 \leq x \leq 1,\\ Kx, \quad x > 1. \end{cases} \end{equation*} Define the sequence $g_n(x) = \frac{1}{n} g(nx)$ and the shifted sequence \begin{equation*} f_n: [0,1] \to \mathbb{R}, \ f_n(x) = g_n(x - 1/2). \end{equation*} Since $g \in C^1[0,1]$, then $f_n \in C^1[0,1]$ for all $n$.
Also, for all $n$, $\Vert f_n' \Vert = \Vert g_n' \Vert = \Vert g'(n(\cdot)) \Vert = K$, since $|4x-3x^2| \leq 1$ for all $0 \leq x \leq 1$. Hence $(f_n)$ is a sequence in $Y$. But $(f_n)$ converges to the function \begin{equation*} f: [0,1] \to \mathbb{R}, \ f(x) = \begin{cases} 0, \quad x \leq 1/2,\\ K(x-1/2), \quad x > 1/2, \end{cases} \end{equation*} which is not in $C^1[0,1]$. Hence $Y$ is not sequentially compact, thus not compact.