= Solution
A <finite CW complex> can be defined without assuming separation properties in advance. Start with a finite discrete set $X^0$. Inductively form $X^k$ from $X^{k-1}\sqcup\coprod_\alpha D^k_\alpha$ by identifying each boundary point with its image under a continuous attaching map $a_\alpha:S^{k-1}\to X^{k-1}$. Only finitely many disks are attached in total, and $X=X^N$ for some finite $N$. The interiors of the disks are its cells. Its <weak topology of a CW complex> is the final topology of the characteristic maps: a subset is closed exactly when its inverse image in every characteristic disk is closed. In this finite construction it is the topology obtained by the successive quotients. Closure-finiteness is automatic because there are only finitely many cells.
Here is a direct proof of both requested topological properties, without presupposing that arbitrary attaching maps give a triangulable space. Suppose the <compact> previous stage $Y$ embeds by $e:Y\hookrightarrow\mathbb R^m$, and attach one $k$-cell, $k\ge1$. Extend the boundary coordinate map radially by
$$
G(ru)=r\,e(a(u)),\quad u\in S^{k-1},\qquad G(0)=0.
$$
It is continuous at zero because the image of the <sphere> is bounded. Map the two pieces into $\mathbb R^{m+k+1}$ by
$$
y\longmapsto(e(y),0,0),\qquad x\longmapsto\bigl(G(x),(1-|x|)x,1-|x|\bigr).
$$
At $|x|=1$ these agree with the attaching identification. They therefore induce a <continuous map> on the quotient. Its restriction to the old stage is injective. An interior point has positive last coordinate $t=1-|x|$, and the preceding coordinate block recovers $x$ by division by $t$. Hence it cannot coincide with an old point or another interior point. The induced map is injective.
The quotient is <compact>, being the image of a <compact> <disjoint union>. A continuous injection into <Euclidean space> is <Hausdorff>, by inverse images of disjoint separating neighborhoods; it is also a <homeomorphism> onto its image because closed subsets of the <compact> source have <compact>, hence closed, images in the <Hausdorff> target. Finite discrete $X^0$ embeds in $\mathbb R$, and induction over the finite list of cells proves \b[$X$ is <Hausdorff> and its weak topology is exactly an induced Euclidean <subspace topology>]. This is <Euclidean embedding by finite cell attachment>; no optimal ambient dimension is needed.
For <Real projective space>, filter by $\mathbb{RP}^0\subset\cdots\subset\mathbb{RP}^n$. The complement of $\mathbb{RP}^{k-1}$ in $\mathbb{RP}^k$ is an affine $k$-cell. A characteristic disk is a hemisphere of $S^k$, whose boundary is attached by the antipodal quotient $S^{k-1}\to\mathbb{RP}^{k-1}$. Thus there is one cell in every dimension $0,\ldots,n$.
The incidence number is the sum of the two sheet contributions. Fixing the orientation of one sheet gives degree one; the other differs by the antipodal map on $S^{k-1}$, whose degree is $(-1)^k$. Consequently the <cellular homology of real projective space> has
$$
C_k=\mathbb Z\quad(0\le k\le n),\qquad d_k=1+(-1)^k=\begin{cases}2&k\text{ even},\\0&k\text{ odd}.\end{cases}
$$
Taking kernels modulo images gives, for $n\ge1$,
$$
\boxed{H_q(\mathbb{RP}^n;\mathbb Z)=\begin{cases}\mathbb Z&q=0,\\\mathbb Z/2&0<q<n\text{ and }q\text{ odd},\\\mathbb Z&q=n\text{ and }n\text{ odd},\\0&\text{otherwise}.\end{cases}}
$$
Over $\mathbb R$, multiplication by two is invertible, so
$$
\boxed{H_q(\mathbb{RP}^n;\mathbb R)=\begin{cases}\mathbb R&q=0,\text{ or }q=n\text{ with }n\text{ odd},\\0&\text{otherwise}.\end{cases}}
$$
Over $\mathbb F_2=\mathbb Z/2$, every differential vanishes, giving \b[$H_q(\mathbb{RP}^n;\mathbb F_2)=\mathbb F_2$ for $0\le q\le n$, and zero otherwise]. For $n=0$ the space is a point and only degree-zero <homology> is present, with the chosen coefficient <group>.
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