We know that $f_n:(0,1)\rightarrow\mathbb{R}$ is a sequence of nondecreasing functions on interval $(0,1)$ and that $f:(0,1)\rightarrow \mathbb{R}$ is a continuous function. Let $A\subset (0,1)$ be a dense subset of $(0,1)$, which fulfills the condition $$ \forall_{x\in A} \ \lim_{n\rightarrow \infty}f_n(x)=f(x).$$ With this assumptions I have to
a) show that $f_n$ converges uniformly to $f$ on every $[a,b]\subset(0,1)$,
b) answer if $f_n$ converges uniformly to $f$ on $(0,1)$.
My attempts:
a) We have to show that $$ \forall_{\epsilon>0} \ \exists_{n_0=n_0(\epsilon)} \forall_{x\in [a,b]}\forall_{n>n_0} \ \ |f(x)-f_n(x)|<\epsilon.$$ For $f_n(x)$, $x\in A$, where $A$ is a dense subset of $(0,1)$ we can observe pointwise convergence, so let's take $[a,b]$ in such a way to make it a dense subset of $(0,1)$. Then for every $\epsilon$ we will be able to choose such $n_0=n_0(\epsilon)$ that $\forall_{x\in [a,b]}\forall_{n>n_0} \ \ |f(x)-f_n(x)|<\epsilon$ (from the point convergence).
b) $f_n$ doesn't converge uniformly to $f$ on $(0,1)$ because on this interval doesn't converge even pointwise.
Are my solutions sensible? If not, I politely ask for help. Thanks in advance and have a good day.
a)
First, we show pointwise convergence. For any $x$ we build increasing $x_k \to x$ and decreasing $x'_k \to x$, belonging to $A$. For any $n$, we have $$f_n(x_k) \le f_n(x) \le f_n(x'_k)$$ Since $f_n$ converges on $A$, $$\forall k\ \forall \epsilon\ \exists n_{k,\epsilon}:\ \forall n > n_{k,\epsilon} \ \ |f_n(x_k) - f(x_k)| < \frac \epsilon 2 \tag{1}\label{eq1}$$ On the other hand, since $f$ is continuous: $$\forall \epsilon\ \exists k_{\epsilon}:\ \forall k \ge k_{\epsilon} \ \ |f(x_k) - f(x)| < \frac \epsilon 2 \tag{2}\label{eq2}$$ (The same for $x'_k$). Therefore, $\forall n > n_{k_\epsilon,\epsilon}$ \begin{alignat}{1} \text{(by \eqref{eq1} for $k \gets k_\epsilon$)} \quad\quad & f(x_{k_\epsilon}) &- \frac \epsilon 2 &\le f_n(x) &\le f(x'_{k_\epsilon}) &+ \frac \epsilon 2 \Rightarrow \\ \text{(by \eqref{eq2} for $k \gets k_\epsilon$)} \quad\quad & f(x) &- \epsilon &\le f_n(x) &\le f(x) &+ \epsilon \end{alignat} I.e., we have a pointwise convergence at point $x$.
Uniform convergence on $[a,b]$ is shown here.
b) It's false: consider $f_n(x) = \begin{cases} -\frac 1x, & x < \frac 1n \\ 0, & x \ge \frac 1n \end{cases}$. For each $x$, we have $f_n(x) \to f(x) = 0$. But convergence is not uniform: for any $n$ and $\epsilon$ there exists $x = \min(\frac 1 {2n}, \frac 1 {2 \epsilon})$: $f_n(x) \le -2 \epsilon$.