Local Connectedness is not Preserved under Continuous Mapping

Theorem
Let $\left({A, \tau_1}\right)$ and $\left({B, \tau_2}\right)$ be topological spaces.

Let $f: A \to B$ be a continuous mapping.

Let $\left({\Img f, \tau_3}\right)$ be the image of $f$ with the subspace topology of $B$.

Let $\left({A, \tau_1}\right)$ be locally connected.

Then it is not necessarily the case that $\left({\Img f, \tau_3}\right)$ is also locally connected.

Proof
Proof by Counterexample:

Let $\left({A, \tau_1}\right)$ be any countable discrete space.

Let $B \subseteq \R$ be the set of all points on $\R$ defined as:
 * $B := \left\{{0}\right\} \cup \left\{{\dfrac 1 n : n \in \Z_{>0}}\right\}$

Let $\left({B, \tau_2}\right)$ be the integer reciprocal space with zero under the usual (Euclidean) topology.

From Discrete Space is Locally Connected, $\left({A, \tau_1}\right)$ is locally connected.

Let $f: A \to B$ be a bijection.

From Mapping from Discrete Space is Continuous, $f: A \to B$ is a continuous mapping.

From Integer Reciprocal Space with Zero is not Locally Connected, $\left({B, \tau_2}\right) = \left({\Img f, \tau_3}\right)$ is not locally connected.