Let us regard $X$ as subspace of $\mathbb C$.
That $X$ and $\overline X$ are not homeomorphic is easy to see: The space $\overline X \setminus \{i\}$ has two path components (proof left as an exercise!), but $X \setminus \{0\}$ has infinitely many path components and for $z \ne 0$ the space $X \setminus \{z\}$ has one path component.
We shall show that the inclusion $i : X \to \overline X$ is a homotopy equivalence.
The idea for constructing a homotopy inverse is this: Map the vertical line $L = \{z \in \mathbb C \mid \text{Re}(z) = 0 \}$ and the "left half circles" $C_n^l = \{ z \mid \lvert z-n \rvert = n, \text{Re}(z) \le n \}$ to $0$ and map the "right half circles" $C_n^r = \{ z \mid \lvert z-n \rvert = n, \text{Re}(z) \ge n \}$ onto the full circles $C_n$ such that the two boundary points $n(1 \pm i)$ of $C^r_n$ are sent to $0$. Let us make this precise.
Define $M = L \cup \bigcup_{n=1}^\infty C_n^l$ and
$$H : M \times I \to \overline X, H(z,t) = \begin{cases} n \left(1 + \dfrac{tz-n}{\lvert tz-n \rvert} \right) & z \in C^l_n \\ tz & z \in L \end{cases} $$
The geometric idea for this definition is this: Given $z \in C^l_n$, consider the line $L(z,t)$ through $n$ and $tz$ (which lies on the secant segment connecting $0$ and $z$). It intersects $C^l_n$ in a single point $H(z,t)$. Note that by construction $H( C_n^l \times I) \subset C_n^l$ and $H(L \times I) \subset L$. Moreover $H(z,0) = 0$ and $H(\overline z,t) = \overline {H(z,t)}$. We shall show later that $H$ is continous.
The map $f_n : [0,\pi] \to C_n, f_n(s) = n(1+e^{is})$, is an embedding whose image is the closed upper half circle. Let $R = \bigcup_{n=1}^\infty C_n^r$. The $C^r_n$ are open subspaces of $R$. Therefore
$$G : R \times I \to \overline X, G(z,t) = \begin{cases} G_+(z,t) = f_n\left(\frac{2f_n^{-1}(z)f_n^{-1}(H(n(1+i),t)}{\pi}\right)) & z \in C^r_n , \text{Im}(z) \ge 0 \\ \overline{G_+(\overline z,t)} & z \in C^r_n , \text{Im}(z) \le 0 \end{cases}$$
is continuous. The geometric idea for this definition is this: Pull the two the boundary points $n(1 \pm i)$ of $C^r_n$ to the left along the circle $C_n$ until they reach the points $H(n(1 \pm i),t)$. This induces a deformation of $C^r_n$ which takes place inside $C_n$.
Since $H, G$ are defined on closed subspaces of $\overline X \times I$ whose union is $\overline X \times I$ and $H, G$ agree on the intersection of these sets, we get a continuous homotopy
$$K : \overline X \times I \to \overline X .$$
Note that $K(X \times I) \subset X$ and $K(z,1) \in X$ for all $z \in \overline X$. Let
$$\rho : \overline X \to X, \rho(z) = K(z,1). $$
Clearly $i \circ \rho$ is homotopic via $K$ to $id$, similarly $\rho \circ i$ is homotopic via $K \mid_{X \times I}$ to $id$.
This proves that $i$ is a homotopy equivalence.
Let is finally verify that $H$ is continuous. This is technically somewhat nasty.
Clearly $H$ is continuous in all points of $(M \setminus L) \times I$ because $M \setminus L = \bigcup_{n=1}^\infty \left(C_n^l \setminus \{0\}\right)$ and the sets $C_n^l \setminus \{0\}$ are open in $M$. Let us show that $H$ is continuous in all points $(\zeta,\tau) \in L \times I$. Since $H \mid_{L \times I}$ is continuous, it suffices to consider a point $(\zeta,\tau) \in L \times I$ and to prove that for each $\epsilon > 0$ one has $\lvert H(z,t) - \tau \zeta \rvert < \epsilon$ for all $(z,t) \in M \setminus L$ which are sufficiently close to $(\zeta,\tau)$. To do so, it clearly suffices to show that
$$\lvert H(z,t) - t z \rvert < \epsilon \text{ if } (z,t) \text{ is sufficiently close to } (\zeta,\tau) .$$
Let $n$ denote the unique index such that $z \in C^l_n$ and write $z = x +iy$. The point $H(z,t)$ was obtained as the intersection of the line $L(z,t)$ with $C^l_n$. For $z \ne n(1 \pm i)$ the line $L(z,t)$ also intersects $L$ and it is easy to see that the intersection point is $g(z,t) = i\dfrac{nty}{n-tx}$. Note that for $z \in C^l_n \setminus \{n(1+i), n(1-i)\}$ we have $tx \le x < n$. Clearly $\lvert H(z,t) - t z \rvert \le \lvert g(z,t) - t z \rvert$. Straightforward calculations show that
$$\lvert g(z,t) - t z \rvert^2 = t^2x^2 + \dfrac{t^4x^2y^2}{(n - tx)^2} \le x^2(1 + \dfrac{y^2}{(n - tx)^2}).$$
Write $\zeta = i\eta$ with $\eta \in \mathbb R$. Then for $\lvert z - \zeta \rvert < \frac{1}{2}$ we have $0 \le tx \le x \le \sqrt{x^2 +(y - \eta)^2} = \lvert z - \zeta \rvert < \frac{1}{2} < n$, thus we are in the situation $z \ne n(1 \pm i)$. Moreover $n - tx \ge n - x \ge 1 - \frac{1}{2} = \frac{1}{2}$ and $\lvert y - \eta \rvert \le \lvert z - \zeta \rvert < \frac{1}{2}$. In particular $\lvert y \rvert < \lvert \eta \rvert + \frac{1}{2}$ and therefore
$$\lvert g(z,t) - t z \rvert^2 \le x^2(1 + 4(\lvert \eta \rvert + \frac{1}{2})^2)$$
which completes the proof because $x \to 0$ as $z \to \zeta$.
Best Answer
In both cases the subspace topology that your union inherits from the plane is metrizable. The corresponding wedge, a quotient of the union, is not. At the vertex the quotient does not have a countable neighbourhood base: given a countable sequence $\langle U_n:n\in\mathbb{N}\rangle$ of neighbourhoods take in the $k$th space a neighbourhood $O_k$ of the point corresponding to the vertex that is a proper subset of $\bigcap_{n\le k}U_n$. Then the $O_k$ determine a neighbourhood $O$ of the vertex that contains none of the $U_n$.