In a lecture, I have written down the following definition for CW complexes.
$X= \bigcup_i \{e_i\}$ and the $\{e_i\}$ form a partition. Furthermore $e_i$ is homeomorphic to $B(x,1)\subset \mathbb{R}^N$ for some $N \in \mathbb{N}$ that depends on the particular $i$. Furthermore there is a continuous extension of this homeomorphism $f_i$ from $f_i: \overline{B(x,1)} \rightarrow X$ satisfying $f_i (\overline{B(x,1)} \backslash B(x,1))\subset X^{\text{dim}(e_i)-1}$(where the latter expression is supposed to denote the respective skeleton). And we also want that $\overline{e_i}$ is contained in a finite union of cells.
What I am particularly sceptical about is this $B(x,1)$ thing, cause everywhere I see people talking about discs and not balls. So, is this the standard definition for a CW complex or is there anything that looks suspicious?
I just added a question concerning this definition:
EDIT: Based on this definition, do I have $f_i( \overline{B(x,1)})= \overline{e_i}$? ($\subset$ is clear, due to continuity, but are they also equal?)
Your definition of a CW complex is missing a couple of conditions: First, you need to assume that $X$ is Hausdorff. Second, you also need to assume that the topology of $X$ is such that a subset $S\subseteq X$ is closed in $X$ if and only if its intersection with each $\overline {e_i}$ is closed in $\overline {e_i}$. (This is traditionally called the "weak topology" associated with the collection $\{\overline {e_i}\}$, and is the reason for the "W" in CW. The "C" stands for "closure finite," which is the condition that each $\overline{e_i}$ is contained in a finite union of cells.)
As for discs vs. balls: The usual terminology in treatments of CW complexes is to say an open cell is a topological space homeomorphic to an open ball in some $\mathbb R^N$, and a closed cell is one that is homeomorphic to a closed ball. A disc is just a $2$-dimensional (open or closed) cell. (The term ball is a little less specific -- it can refer to an open or closed ball in any metric space, which is why the "cell" terminology is preferable.)
To answer your last question: Yes, it is true that $f_i(\overline{B(x,1)})= \overline{e_i}$. The inclusion $\supseteq$ follows from the fact that $\overline{B(x,1)}$ is compact, and therefore its image is compact and thus closed in $X$. (This is one reason why the Hausdorff condition is needed.)