This is related to 6.3.24 in Engelking's Topology book.
The Hilbert space $H$ is the set of sequences $(x_i) \in \mathbb R ^\omega$ such that $\|x\|=\sum _{i=1} ^\infty x_i ^2<\infty$, with metric $d(x,y)=\sqrt{\sum_{i=1} ^\infty (x_i-y_i)^2}$.
Let $X$ be the subspace of $H$ consisting of the sequences of rationals.
Observe that by assigning a point $(x_i)\in X$ with the same point in $\mathbb Q ^\omega$ one defines a continuous mapping and deduce that there is a one-to-one continuous mapping $f:X\to C$ of $X$ into the Cantor set $C$ such that $f(x_0)=0$.
By the observation we may assume $X$ is second countable and zero dimensional. Let $x_0\in X$. Let $\{U_i:i\in\omega\}$ be a clopen base for $X$. Let $A=\{i\in\omega:x_0\in U_i\}$ and $B=\omega\setminus A$. For each $x\in X$ let
$$f(x)(i)=\begin{cases} 0 &\mbox{if } (i\in A \wedge x\in U_i) \vee (i\in B \wedge x\in X\setminus U_i) \\ 1 & \mbox{else} \end{cases}.$$
Then $f:X\to 2^\omega$ by $f(x)=(f(x)(i))$ is well-defined, one-to-one, continuous, and $f(x_0)=0$.
Define a mapping $g:X\to [0,1]^2$ by letting $g(x)= \big( \frac{f(x)}{\max (1,\|x\|)},\frac{\|x\|}{1+\|x\|} \big)$. Show that the space $X_1=g[X]\cup \{(0,1/2)\}$ is hereditarily disconnected but not totally disconnected. Hint: Note that there is a one-to-one continuous mapping of $g[C]$ onto $f[X]$, and that if $A\subseteq X$ is clopen and $\{\|x\|:x\in A\}$ is bounded then $A=\varnothing$.
I assume that $g[C]$ should be $g[X]$.
$X_1$ is hereditarily disconnected: For $(r,s)\in g[X]$ let
$$h(r,s)=\begin{cases} r &\mbox{if } s\leq 1/2 \\ {\frac{rs}{1-s}} & \mbox{else} \end{cases}.$$
It is easy to show that $h$ is a continuous one-to-one mapping of $g[X]$ onto $f[X]$. As $f[X]$ is hereditarily disconnected, so is $g[X]$. Also, in creating this mapping I realized we can think of $g[X]$ as a subset of $C\times [0,1]$, squeezed at the top to the point $(0,1)$. It is now clear that no connected subset of $X_1$ meets $\{(0,1/2)\}$ and $g[X]$, as $(0,0)$ is the only point of $g[X]$ along the $y$-axis.
$X_1$ is not totally disconnected: Suppose $A$ is clopen in $X_1$ containing $(0,1/2)$, missing $(0,0)$. There exist $\epsilon,\delta>0$ such that $\big(B_\epsilon (0)\times B_\delta (1/2)\big)\cap X_1\subseteq A$. There exists $r\in (0,\epsilon)$ such that $[0,r]\cap C$ is clopen in $C$. Then I BELIEVE $$\big([0,r]\times [0,1/2]\big)\cap g[X]\setminus A$$ is clopen in $g[X]$. It is also nonempty as it contains $(0,0)$. Its preimage under $g$ is nonempty and clopen in $X$, and has bounded norm. This contradicts the claim in the 2nd part of the hint, which is provable via the standard argument that $X$ (the Erdos space) is not zero dimensional.
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NOTE1: I believe I have solved this problem now, but I would of course appreciate any feedback/verification. Also, if anyone is interested I could add some details to my solution.
NOTE2: For the purpose of giving a space which is hereditarily disconnected but not totally disconnected, I think we could have defined $g$ simply by $g(x)=(f(x),\|x\|)$, and let $X_1=g[X]\cup \{(0,r)\}$ for any $r\in (0,\infty)$. The definition above is more complicated in order to achieve the squeezing, so that adding $(0,1)$ to the space actually makes it connected (by the 2nd part of the hint). That is the subject of another part of the exercise.