Closure of a set of functions (w.r.t. $\infty$-norm)

Question

I want to show that a certain set of functions ($\{f\in\mathcal{B}([0,1],\mathbb{R}):f(0)=0, f \text{ is continuous at } 0\}$, the set of bounded functions on the interval $[0,1]$ with the given properties) is closed in the set of all bounded functions on the interval $[0,1]$, with respect to the infinity norm (on the interval $[0,1]$). The definitions that I am using are:
$\bar{V}:=\{f\in X : d(f,V)=0\}$, where $d(f,V)=\inf\{||f-g||_\infty:g\in V\}$ and $||f||_\infty=\sup_{x\in[0,1]}|f(x)|$.
My question regarding this is the following: if $d(f,V)=0$, then we have $$\inf\{||f-g||_\infty : g\in V \}=0\implies \inf\{\sup_{x\in[0,1]}|f(x)-g(x)|:g\in V\}=0$$
which means that the difference between the functions on the specified interval is at most $0$, so $f=g\in V$. If this is indeed true, then the proof was suspiciously easy. However, I feel like I made a mistake going from the definition to the implication. Did I substitute the norm in the correct manner? It seems like feels like what I produced here is a tautology, more so than a proof, but I can't tell what the flaw is.

Answer 1

Consider this example.

Let $C[0,1]$ be the vector space. Let $V=\{f_n\in C[0,1]: f_n(x)=1/n \ \forall n\in\mathbb Z^+\}$.

If we set $f(x)=0, \ \forall x\in [0,1]$, then $f\in \overline V$. But $f\neq f_n$ for any $n$. So $f\notin V$.

Answer 2

Take $f_n(x) = 1-(1-x)^n$. The limit using the above norm is $f=1_{(0,1]}$ which is not continuous at $0$.