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With the quotient realization defined in part (i), the answers are no and no. Write $W=|K|$ as countably many closed arms $[0,1{]}_n$ with all their zero endpoints identified to $v$.

First, the outer vertices $D=\{u_n:n\ge 1\}$ form a closed discrete subset. Its intersection with every closed arm is the closed singleton at coordinate one, so $D$ is closed by the weak topology. The terminal half-arm $(1/2,1{]}_n$ is an open neighbourhood of $u_n$ meeting $D$ only there. Thus $D$ is infinite, closed and discrete. Since closed discrete subsets of compact spaces are finite, $W$ is not compact. A homeomorphism to a compact subset of the plane would make $W$ compact, which is impossible.

For the stronger obstruction, suppose $(U_n)$ were a countable neighbourhood basis at $v$. Each $U_n$ contains an initial interval in arm $n$, so choose $x_n\in U_n$ on that arm with coordinate $t_n\in (0,1)$. Define $V$ to meet arm $n$ in $[0,t_n/2)$. It is an open neighbourhood of $v$, because each of these intersections is open in the closed interval $[0,1]$. But $x_n\in U_n\setminus V$ for every $n$, so no $U_n$ is contained in $V$. This contradicts the basis condition.

Hence $W$ is not first countable. Every metric space is first-countable, with local basis of balls of radius $1/n$, and this property passes to subsets with the subspace topology. Thus $W$ is not homeomorphic to any subset of ${\mathbb{R}}^2$. This is the weak realization of a countable star.

The realization convention matters. For the metric realization of a simplicial complex using Euclidean distance on barycentric coordinates, the second answer would be yes: map the point on arm $n$ at parameter $t$ to $t{e}^{i/n}$. Each ray is isolated away from zero, and at zero both spaces have neighbourhood bases given by a single radius on all arms. This gives the metric realization of a countable star as a planar subset, still noncompact. It does not give the quotient topology used in part (i).