Contractibility of convex set Suppose that $\Omega$ is a convex open subset of an infinite dimensional vector space $E$ such that $\Omega$ is not contained in any finite dimensional subspace of $E$.
Let $Q_m\subset \Omega$ denote a set of $m\in \mathbb{N}_{>0}$ distinct points.  

Question:
  Is the space $\Omega\setminus Q_m$ contractible?

Added later:
Assume that for any set $Q_m'$ of $m$ distinct points $E\setminus Q'_m$ is contractible (I believe that this is true, see my answer below).
By "blowing up $\Omega$ like a balloon" we get a homeomorphism $\varphi$ from the convex open set $\Omega$ to $E$.
More precisely, fix $x_0\in\Omega$ and for any open interval $I\subset \mathbb{R}$ let $\varphi_I:I\to \mathbb{R}$ be a (suitable) homeomorphism.
Now, define $\varphi:\Omega \to E$ by $$\varphi(x):=\begin{cases}\varphi_{\Omega\cap \mathbb{R}\cdot (x-x_0)}(x), \text{ if } x\neq x_0 \\ x_0 , \text{ if } x= x_0.\end{cases}$$
(I know that the definition of $\varphi$ is not formally correct but I think that the idea is clear.)
Restricting $\varphi$ to $\Omega\setminus Q_m$ we get a homeomorphism between $\Omega\setminus Q_m$ and $E\setminus \varphi(Q_m)$.
By assumption $E\setminus \varphi(Q_m)$ is contractible and hence $\Omega\setminus Q_m$ is also contractible.
Is the intuition behind this heuristic argument any good?
Can it be made into a complete and rigorous argument?
 A: Following the suggestion of Martin I'll post an answer to my own question. I would appreciate some feedback. Please let me know if you notice any errors.
Let $Q'_m\subset E$ be a set of $m\in \mathbb{N}$ distinct points.

Claim: $E\setminus Q'_m$ is contractible.

Proof:
Write $E$ as $E\cong W\oplus V$ where $W\subset E$ is a finite dimensional subspace containing $Q'_m$.
Let $\Phi:W\oplus V \to W\oplus V$ be a linear map such that $\Phi|_W=id_W$ and $\Phi|_V$ has no non-zero eigenvalues. (For example $\Phi|_V$ could be chosen as $(v_1,v_2,v_3,\ldots)\mapsto (0,v_1,v_2,v_3,\ldots)$.)
The map $\Phi$ gives rise to an isomorphism from $E$ to a genuine subspace $im(\Phi)\subsetneq E$.
It also induces a map $\Phi:E\setminus Q'_m\to E\setminus Q'_m$.
Now, $\Phi_t:E\setminus Q'_m\to E\setminus Q'_m$ defined by
$$\Phi_t(x):=(1-t)x+t\Phi(x)\quad t\in [0,1]$$
is a homotopy from $id_{E\setminus Q'_m}$ to $\Phi$.
Pick $x_0\notin im(\Phi)\cup Q'_m$.
Then $\Psi_t:E\setminus Q'_m\to E\setminus Q'_m$ given by
$$ \Psi_t(x):=(1-t)\Phi(x)+tx_0\quad t\in [0,1]$$
is a homotopy from $\Psi_0=\Phi$ to the constant map $\Psi_1\equiv x_0$.
Hence, $E\setminus Q'_m$ is contractible.$\qquad\square$
This claim together with the fact that $\Omega\setminus Q_m$ is homeomorphic to $E\setminus Q'_m$ (see the comments by Martin and my argument in the question) shows that $\Omega\setminus Q_m$ is contractible.
I'd like to thank Martin for his helpful comments.
