Construction of an infinitely Fréchet differentiable function with given set of zeros in a Banach space After looking at this question, I am now wondering if the following is true.

Let $X$ be a separable Banach space over $\mathbb R$ or $\mathbb C$, and $A\subseteq X$ a closed set. Then there exists a function $f: X\to\mathbb R$ (Cannot be $\mathbb C$ here; always $\mathbb R$ even if the previous one is $\mathbb C$) such that $A=\{x\in X:f(x)=0\}$ and $f$ is infinitely differentiable in the sense of Fréchet (see Fréchet derivative).

Or can it be true after some minor modifications?
Here is my attempt to prove it.
Let $\phi:\mathbb R\to[0,\infty)$ be a $C^\infty$ function satisfying $\phi(t)=1$ for $|t|\leq1/2$ and $\phi(t)=0$ for $|t|\geq 1$. Such $\phi$ can be constructed by using functions like $\exp(-1/t)$.
Since $X$ is separable and $X\backslash A$ is open, $X\backslash A$ can be expressed as a countable union of open balls,
$$
X\backslash A=\bigcup_{k=1}^\infty B(x_k,r_k).
$$
Define, for $x\in X$,
$$
f_k(x)=\phi\left(\frac{\|x_k-x\|}{r_k}\right).
$$
Each $f_k$ is infinitely differentiable, since for $x\neq x_k$, this follows from the chain rule, and if $x=x_k$, obviously all orders of derivatives are zero since it is constant in a neighborhood of it.
Now, define $f(x)=\sum_k a_k f_k(x)$, where
$$
a_k=\frac{1}{2^k} \left(\sup_{x\in B(x_k,r_k)\\ m\le k}\|D^m f_k(x)\|\right)^{-1},
$$
any if the thing in the bracket is zero, we can choose any positive value for $a_k$ as we like.
Does my proof work? 
 A: In a few words, I believe there is not much to add to you proof to make it work, but it nevertheless highly decreases the breadth of the result.
Edit: As Pietro Majer said in the comment, the theorem below actually hold without the uniform bound for the derivatives. In other words, every open set of a fixed Banach space $B$ is the zero set of a $\mathcal C^k$ function if and only if there exists a $\mathcal C^k$ bump function, i.e. a non-zero function with bounded support. A self-contained reference is R. Bonic and J. Frampton, Smooth Functions on Banach Manifolds, 1966. The proof of Theorem 1 contains everything needed to prove the result I state here, up to a notational typo (maybe I am the only one to be terribly confused); the definition of $V_x$ should read:
$$ V_x = \{y:f_{n(x)}(y)>\alpha(x)\}. $$
Notations
I write $L^k(B;B')$ for the Banach space of continuous $k$-linear functions with arguments in $B$ and image in $B'$; also $L(B,B')$ will mean $L^1(B;B')$. By definition, I say that $f:B\to B'$ is differentiable at $x$ only if there exists $A\in L(B,B')$ such that for any $\varepsilon>0$, we have
$$ |f(x+h) - f(x) - A(h)|_{B'} \leq \varepsilon|h|_B $$
for all $h$ small enough. In particular, I always ask for the derivative at some point to be continuous. I define the $\mathcal C^k$ classes as usual, and the higher derivatives $D^k\!f$ of a function $f$ are functions from $B$ to $L^k(B;B')$.
Recall the following classic fact from metric topology.

Fact.
Let $B$ and $B'$ be two Banach spaces. The space of bounded continuous functions $f:B\to B'$ is complete, with respect to the norm
  $$ |f|_{B\to B'} := \sup_{x\in B}|f(x)|_{B'}. $$

It is actually true for $B$ a topological space, and $B'$ a complete metric space.
Positive results
I will show, using your proof, the following result.

Theorem.
Let $B$ be a separable Banach space and $0\leq k\leq\infty$. Suppose that there exists some non-zero function $f:B\to\mathbb R$ of class $\mathcal C^k$ with bounded support, such that
  $$ \sup_{x\in B}|D^\ell\!f(x)|_{L^\ell(B;\mathbb R)}<\infty $$
  for all $0\leq\ell\leq k$. Then any closed set of $B$ is the zero set of some function of class $\mathcal C^k$.

The requirement seems absurd from a finite-dimensional perspective, but actually two things can go bad in infinite dimensions. First off, it might not be easy to find smooth functions to begin with — I am not an expert, but I seem to recall smooth functions with bounded support are not a given in an arbitrary Banach space. Second, a smooth function with bounded support can actually be unbounded. I give an example at the end of this answer.
Before giving the proof, here is a proof of existence in a particular case.

Proposition.
If $H$ is a Hilbert space, the squared norm $x\mapsto |x|_H^2$ is smooth, and all its derivatives are bounded on bounded sets.

Of course $|x|^2_H$ is continuous and quadratic, so its derivatives are $|x|_H^2$ (of order 0), $h\mapsto2\langle x,h\rangle$, $h,h'\mapsto2\langle h,h'\rangle$ and zero (of order $k\geq3$), and we see that they are bounded on bounded sets. Hence for instance $\phi(|x|_H^2)$, with $\phi$ as in your question, will work for the above theorem (we can choose $k=\infty$).
Proof
The proof of the Theorem is basically the one you gave, where you implicitly proceeded by induction using the following elementary lemmas.

Lemma 1.
Suppose $f_n:B\to B'$ is a sequence of functions of class $\mathcal C^1$. Assume moreover that $f_n$ and $Df_n:B\to L(B,B')$ are Cauchy sequences with respect to $|\cdot|_{B\to B'}$ and $|\cdot|_{B\to L(B,B')}$ respectively.
Then $f_n$ and $Df_n$ converge to some continuous functions $f$ and $g$, and $f$ is $\mathcal C^1$ with $Df=g$.

$~$

Lemma 2.
  Let $B$ be a separable Banach space, $U_0$ a bounded open neighbourhood of $0$ in $B$, and $(x_n)_{n\geq 0}$ a dense sequence in $B$.
Then for any open subset $U$ of $B$ and any $x\in B$, $x+r(x)U_0\subset U$ for
  $$0<r(x):=\sup\left\{r>0\text{ such that }x+r U_0\subset U\right\},$$
  and $U=\bigcup_{n\geq0}\big(x_n+r(x_0)U_0\big)$.

To prove the first result, you just have to use the above Fact, and take the limit in
$$ f(x+h) = f(x) + \int_0^1Df(x+th)(h)\mathrm dt. $$
The second one is easy, and I imagine you know it already.
A ‘negative’ result
I am giving here an explicit example of a smooth function $B\to\mathbb R$ with bounded support, but which is itself unbounded.
Suppose you have a continuous inclusion of Hilbert spaces $H\hookrightarrow K$. Then $|x|_K\leq C|x|_H$ for $C>0$ large enough. However, maybe $|x|_K$ can be arbitrary small for $|x|_H=1$. Both $|x|_H^2$ and $|x|_K^2$ are smooth functions on $H$, as discussed in the Proposition, so
$$ f:x\mapsto \frac{\psi\big(1/|x|_H^2\big)}{|x|_K^2} $$
is smooth provided $\psi:\mathbb R\to\mathbb R$ is smooth and vanishes on a neighbourhood of zero.
Now if $x_n$ is a sequence in $H$ such that $|x_n|_H=1$ but $|x_n|_K$ goes to zero, fix some $\psi$ with support in $(1/2,2)$ and $\psi(1)=1$. Then for $f$ defined as above, $f(x_n)$ is $1/|x_n|_K^2$ and diverges, even though $f$ is smooth and has support in the ball of radius $2$ centred at zero.
For an actual concrete example, take the inclusion of Lebesgue spaces $\mathrm L^2\big([0,1],\mathrm dx\big)\subset\mathrm L^2\big([0,1],x\mathrm dx\big)$ and
$$x_n:t\mapsto\begin{cases}
n\text{ if }t\leq 1/n^2,\\
0\text{ else.}\end{cases}$$
