Let $(\mathcal{F}_0,\Vert\cdot\Vert_\infty)$ be a metric space of continuous functions that is dense in the space $\mathcal{F}$. Is the following statement correct:
For every $f\in\mathcal{F}$, there exists a sequence $(f_n)_{n\in\mathbb{N}}$ in $\mathcal{F}_0$ that converges uniformly to $f$.
Specifically, I wonder if we can demand uniform convergence?
Thanks @Qiaochu Yuan and @Dave L. Renfro for the helpful advice. From what I have learnt, I think I can answer my question as follows:
Assume that $(f_n)_{n\in\mathbb{N}}$ converges to $f$ with respect to $\Vert\cdot\Vert_\infty$, that is, for all $\varepsilon > 0$, there exists $N\in\mathbb{N}$ such that for all $n > N$ \begin{equation} \Vert f_n - f\Vert_\infty = \sup_{x\in\mathcal{X}} \vert f_n(x) - f(x)\vert < \varepsilon \end{equation} Hence, \begin{equation} \forall\,\varepsilon > 0, \exists\,N\in\mathbb{N} : \forall\,n > N, x\in\mathcal{X}\qquad\vert f_n(x) - f(x)\vert < \varepsilon \end{equation} which coincides with the definition of uniform convergence. To conclude, we realize that $\mathcal{F}_0$ being dense in $\mathcal{F}$ and convergence in the sup-norm imply uniform convergence of $(f_n)_{n\in\mathbb{N}}\subset\mathcal{F}_0$ to $f\in\mathcal{F}$.