Let $X$ be a finite CW-complex. Then there is an embedding $X\to\mathbb{R}^n$ for some $n$. To show this it suffices to consider the case when $X$ is obtained from a CW-complex $Y\subset B^m\subset \mathbb{R}^m$ by attaching a cell: $X=Y\cup_f B^k$ where $B^m$ is the unit ball in $\mathbb{R}^m$ and $f:S^{k-1}\to Y$ is the attaching map.

Let us show that there is an embedding $i:B^m\sqcup B^k\to \mathbb{R}^n$ for some $n$ so that, given $x',x''\in B^m$ and $y',y''\in B^k$, the segments $[i(x'),i(y')]$ and $[i(x''),i(y'')]$ do not intersect unless $x=x''$ or $y'=y''$. Indeed, let's assume both balls live in some $\mathbb{R}^r$ and take $\mathbb{R}^n$ to be the space of polynomials $\mathbb{R}^r\to\mathbb{R}$ of sufficiently high degree (4 will do). Then for any 4 points in $\mathbb{R}^r$ the conditions that an element of $V$ are zero at those points are linearly independent. So for any $x\in B^m\sqcup B^k$ take $i(x)\in V^*$ to be the evaluation function $p\mapsto p(x)$.

Now set $X'$ to be the union of $i(Y), i(B^k)$ and all segments $[i(x),i(f(x))]$ for $x\in S^{k-1}$. there is a natural bijective continuous map $X\to X'$, which is a homeomorphism since $X$ is compact and $\mathbb{R}^n$ is Hausdorff.

Remarks:

From this construction it is clear that one can construct an embedding of $X$ that is smooth on the interior of each cell. So by slightly modifying the proof of Whitney's embedding theorem we see that $X$ can be embedded in $\mathbb{R}^{2\dim X+1}$. However, the strong Whitney theorem does not hold. E.g. there are 2-polyhedra that can't be embedded in $\mathbb{R}^4$, see e.g. theorem 3 in this paper http://www.fmf.uni-lj.si/~repovs/clanki/2001/OnReSk01.pdf

If $X$ is not assumed finite then I would guess that a necessary and sufficient condition for $X$ to be properly embeddable in some $\mathbb{R}^n$ is that $X$ should be locally finite and should have at most countably many cells.