Prove that $X := \{ f: [0,1] \to [0,1] : f \text{ is continuous and } f(1) = 0 \}$ with the given distance is neither connected nor separable 
Prove that $X := \{ f: [0,1] \to [0,1] : f \text{ is continuous and } f(1) = 0 \}$ with $d(f,g) := \inf \{r \geq 0 : f(t) = g(t) \forall r ≤ t ≤ 1 \}$ is neither connected nor separable.

Here is my work so far: let $y \in [0,1]$ and consider the function $f_y \in X$ that is $y$ on $[0,99/100]$ and is linear and continuous on $[99/100 , 1]$. By construction, $f_y \in X$ for all $y \in [0,1]$. If we assume that $X$ is separable, then there exists a countable dense subset $\\{g_n\\}$, so that for any given $y \in [0,1]$ and $\epsilon > 0$ there exists some $g_n$ such that $d(f_y,g_n) < \epsilon$, so that $f_y(x) = g_n(x)$ for all $x \in [\epsilon,1]$. By looking at the values $g_n(98/100)$, we see that $\\{ g_n(98/100) \\} = [0,1]$, a contradiction.
Does this work look okay? I was also am lost on how to prove that $X$ is disconnected. I cannot think of a disconnection of $X$.
 A: For $a>0$ let $f_a(x)=a(1-x).$ Then $$d(f_a,f_b)=1,\qquad a\neq b$$
Therefore the space is not separable.
Concerning connectedness, any closed ball with radius less than $1$ is open. As the space $X$ is linear it suffices to consider the balls centered at $0.$  Let $$S_r=\left \{f\in X\,:\, d(f,0)\le r\right\},\quad 0<r<1.$$
Clearly $S_r\neq X,$ as $d(1-x,0)=1.$
The set $S_r$ is closed by definition. It is open, as well. Indeed, assume $f\in S_r.$ Then
$f(x)=0$ for $x\in [r,1].$  Let
$d(f,g)<r,$ Then $f(x)=g(x)$ for $[r,1].$ In particular $g(x)=0$ for $[r,1].$ Hence $d(g,0)\le r,$ i.e. $g\in S_r.$ Thus $S_r$ is an open subset of $X.$
Remark A similar proof shows that any sphere with radius less than $1$ is open.
A: The space $X$ is indeed disconnected. Any property that is "local" around $x=1$ will give an open and closed subset. For example $$ S = \{ f \in X \ : \ f \text{ is differentiable in a nbhd of $x=1$} \} $$ is both open and close (clearly $\emptyset \neq S \neq X$). Pick any function $f\in S$, then all functions $g\in B_{1/2}(f)$ coincide with $f$ in a nbhd of $x=1$, so clearly all those functions are again differentiable in a nbhd of $x=1$ and so $g\in S$. It is also closed, because if $(f_n)_n \subseteq S$ converges to $f$, then at some point $d(f_n,f)<1$ and hence, $f$ coincides with some $f_n$ in a nbhd of $x=1$ is therefore differentiable in a nbhd of $x=1$, i.e. $f\in S$.
For separability I'd phrase it slightly differently (though in spirit it is the same as yours). Assume it is separable and that $(g_n)_{n\in \mathbb{N}}$ is dense in X. As the family is countable, so is the set $\{g_n(1/2) : n\in \mathbb{N}\}\subseteq[0,1]$. Hence, there exists some value $y\in [0,1]\setminus \{ g_n(1/2) \ : \ n\in\mathbb{N}\}$ and thus, $g_n(1/2)\neq y=f_y(1/2)$ (here I choose the same $f_y$ as in the OP) which implies $d(f_y,g_n) \geq 1/2$, which is a contradiction to the assumption that $(g_n)_{n\in \mathbb{N}}$ is dense in $X$.
