Let $M$ = subset of $\mathbb{R}$. Let sequence of functions $f_n(x)$ converges uniformly to $f(x)$ defined on $M$. Now we choose nonempty $N$ = subset of $M$. Then $f_n(x)$ also converges uniformly to $f(x)$ on $N$.
My solution:
Uniform convergence definition $(\forall \varepsilon >0)(\exists a_0\in A):a\in A\wedge a\ge a_0\wedge x\in M\Rightarrow |f_a(x)-f(x)|<\varepsilon $
We have $N$ = subset of $M$ then $x$ above is $x\in N$, so it holds: $(\forall \varepsilon >0)(\exists a_0\in A):a\in A\wedge a\ge a_0\wedge x\in N\Rightarrow |f_a(x)-f(x)|<\varepsilon $