$\mathbf {The \ Problem \ is}:$ If $X$ is a countable union of algebraic subsets of $\mathbb C^n$, i.e. a set defined by algebraic equations in the coordinates of $\mathbb C^n$(under standard topology) then show that $P= \mathbb C^n \setminus X$ is poly-line connected i.e. whose any two points can be joined by finitely many broken lines lying in that set .
$\mathbf {My \ approach} :$ Actually,I tried in this way that for any two point $p,q \in P$, we join them by a line $L$, the line segment $L_{pq}$ is connected and as $X$ is closed, we let two open balls $N_p$ and $N_q$ of $p$ and $q$ such that both of them doesn't meet $X .$
Then for any point on $L_{pq}$, not in $X$, we let such a neighborhood not meeting $X$ and for points in $X$, we let an $\epsilon$ neighborhood, they cover the connected set $L_{pq}$, then I think I need to use the finite chain criterion of connected sets and poly-line connectedness of $\mathbb C^n \setminus K$ where $K$ is countable .
But, I can't think of the fact where to use the "algebraic set" .
I found a result that such sets are nowhere dense in $\mathbb R^n .$
Best Answer
This is an edition of my answer, which I found after rereading pretty poorly motivated.
Let $X=\bigcup_k X_k$, where each $X_k\subset\mathbb C^n$ is a complex algebraic subset. For any $p,q\in P$ the complex line $L_{pq}$ is not contained in $X_k$, hence $X_k\cap L_{pq}$ is a proper algebraic subset of $L_{pq}$, hence a finite set of points. Thus $X\cap L_{pq}$ is countable. Now, the complex line $L_{pq}\subset\mathbb C^n$ is a real plane $H\subset\mathbb R^m$, $m=2n$, and we can repeat an argument given somewhere in mathstackexchange: pick a real line $\ell\subset H$ through $p$ that does not meet $H\cap X$ (possible because this set is countable); next pick another $\ell'\subset H$ through $q$ that does not meet $H\cap X$ and is not parallel to $\ell$. Then the intersection point $x\in\ell\cap\ell'$ gives a polygonal path $\gamma$ from $p$ to $x$ in $\ell$, then from $x$ to $q$ in $\ell'$, which does not meet $X$. We are done.
This is an instance of a more general fact: if $X\subset\mathbb R^m$ is a countable union of smooth submanifolds $M_i\subset\mathbb R^m$ of codimension $\ge2$, then $P=\mathbb R^m\setminus X$ is connected by polygonal paths.
Indeed, every complex algebraic set $X_k$ in $X=\bigcup_k X_k$ is a union of complex analytic submanifolds of ${\mathbb C}^n\equiv{\mathbb R}^m$, which have an underlying structure of real smooth submanifolds. Namely, regular locus, then regular locus of the singular locus, then regular locus of the singular-singular locus... The topological real codimension of those manifolds is at least $2$. Thus we have that $X\subset{\mathbb R}^m$ is a countable union of smooth manifolds of codimension at least $2$. In passing, note this works the same for complex analytic subsets $X_k$ of ${\mathbb C}^n$, each one being a countable union of complex analytic submanifolds.
Now let us see that $P=\mathbb R^m\setminus X$ is connected by polygonal paths. Pick any two points $p,q\in Y$. Then every plane $H$ containing both points is given by a third point $c\in\mathbb R^m$, and by an standard transversality argument, the $c$ can be chosen for $H$ to be in general position with all $M_i$. Indeed, let $r$ be the real line through $p,q$. The plane $H$ generated by $r$ and a third point $c\notin r$ can be parametrized by $(1-s-t)c+sp+tq$, and we consider the smooth mapping $$ F:(\mathbb R^m\setminus r)\times(\mathbb R^2\setminus\{s+t=1\})\to\mathbb R^m $$ given by $$ F(c,s,t)=F_c(s,t)=(1-s-t)c+sp+tq. $$ We exclude $s+t=1$ so that all partial derivatives $\partial F/\partial c=(1-s-t)$Id with $1-s-t\ne0$ are linear isomorphisms, which guarantees that for a residual set $C_i$ of $c$'s the partial mapping $F_c$ is transversal to $M_i$ (parametrized version of density of transversality). This implies the inverse image $F_c^{-1}(M_i)\subset\mathbb R^2\setminus\{s+t=1\}$ is a manifold of codimension equal to that of $M_i$, hence at least $2$. Consequently $Y_i=F_c^{-1}(M_i)$ has dimension at most $0$ and it is a countable set. As $F_c(Y_i)=M_i\cap H\setminus r$, we see that $M_i\cap H\setminus r$ is countable. In other words, $H$ is in general position with $M_i$.
Thus $C=\bigcap_iC_i$ is residual, hence dense, hence not empty, and any $c\in C$ gives a plane $H$ transversal to all $M_i$'s. Thus all intersections $M_i\cap H\setminus r$ are countable, and $X\cap H\setminus r$ is countable. We can produce a polygonal path in $H$ from $p$ to $q$ using two lines $\ell$ and $\ell'$ as above, with the additional precaution that they are chosen different form $r$ to avoid the posible intersection $X\cap r$.