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In the metric realization of a simplicial complex, the point on arm $n$ at parameter $t$ is $(1-t)e_v+t{e}_{u_n}$. Its distance from the common vertex is $\sqrt{2}t$. Consequently the vertex has a neighbourhood basis consisting of one uniformly short initial interval on every arm. Away from that vertex, each arm is locally isolated from the other arms.

The map $(n,t)\mapsto t{e}^{i/n}$, with all $t=0$ identified, is a homeomorphism ontoIt is a bijection, preserves the one-dimensional topology on each arm away from zero, and matches the uniform-radius neighbourhood bases at zero. The positive real limiting ray is omitted. The outer endpoints form an infinite closed discrete subset in the original metric realization, so the image is not compact. This planar fan and the weak realization of a countable star have different topologies at their common vertex.

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).