A continuous image of a second countable space which is not second countable.

Take $f: (\mathbb R, T_{Euclid}) \to (\mathbb R, T_{cof})$ the identity map from the standard topology to the cofinite topology on $\mathbb R$.

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To see why a basis for the cofinite topology on $\mathbb R$ cannot be countable (by contradiction) assume that you have a countable basis $B$ of $T_{cof}$. Then for every $O$ in $B$, $O^c = \mathbb R \setminus O$ is finite (by definition). Take the union $\bigcup_{O \in B} O^c$ over all $O^c$ in $B$. Then $\bigcup O^c$ is countable hence a proper subset of $\mathbb R$. Pick a point $x$ in $\mathbb R \setminus \bigcup O^c$. Then $\mathbb R \setminus \{x\}$ is open in the cofinite topology but you cannot write it as the union of sets in $B$. Hence $B$ cannot be a basis.

How about a separable Banach space $X$, with $\tau$ the norm topology and $\tau_w$ the weak topology. The identity map $(X, \tau) \to (X, \tau_w)$ is continuous, and $(X,\tau)$ is a separable metric space, hence second countable, but $(X, \tau_w)$ is not even first countable.

This is exercise 16B.1 in Willard's General Topology: For each $n\in \Bbb N$, let $I_n$ be a copy of $[0,1]$. Let $X$ be the disjoint union of the spaces $I_n$. Identify the left hand endpoints of all the $I_n$ and let $Z$ be the resulting quotient space. The distinguished point in $Z$ has no countable nhood base, though $X$ is second countable.