Prove $A_m$ is closed in $C[0,1]$

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Let $m\in\mathbb{N}$ arbitrary and $A_m \subset C[0,1]$, the subset of all function $f$ verify: Exists $x_0\in[0,1-\frac{1}{m}]$ such that: $\frac{f(x_0+h-f(x_0)}{h}\leq m \forall h\in(0,\frac{1}{m}$

Prove $A_m$ is closed in $C[0,1]$

My idea is:

Prove $A_m$ is bounded uniformly and equicontinuous then by Ascoli theorem $A_m$ be compact.

This implies $A_m$ is closed set.

But i don't know how to prove this, can someone help me?

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Fix $h$. $f\to \inf_{x\in [0,1-\frac 1 m]} \frac {f(x+h)-f(x)} h $ is a continuous function on $C[0,1]$. This is a fairly strightforward verification. [Let me know if you need help with this]. Hence $\{f\in C[0,1]:\inf_{x\in [0,1-\frac 1 m]} \frac {f(x+h)-f(x)} h \leq m$ is a closed set. Take intersection over $h$ to see that $A_m$ is closed.