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 $X\cap r$.
Best Answer
Any path connected space is connected. Show that the two sets $\{ (x, y) : y \gt 0 \}$ and $\{(x,y) : y \lt 0 \}$, are disjoint open sets whose union is the whole space. That is the definition of a disconnected space. Since it's not connected, it can't be path connected, since that would lead to a contradiction.
To show formally that any line dividing $\mathbb{R}^2$ results in a disconnected space. Consider the homeomorphism $z = x + yi \mapsto e^{i\theta}z + z_0$. It maps lines to any other line given the angle to the new line and the proper offset. Since $\mathbb{R}^2$ is homeomorphic to $\mathbb{C}$ you have a sequence of homeomorphisms from your cut space to a cut space with an arbitrary line, so any line cutting $\mathbb{R}^2$ results in a disconnected space.