For a map of metric spaces the following agree: $\varepsilon$-$\delta$ continuity everywhere, preimages of open sets are open, preimages of closed sets are closed, sequential continuity, and $f(\overline{A}) \subseteq \overline{f(A)}$

(a) implies (b): let $V \subseteq Y$ be open and $x \in f^{-1}[V]$; since $f(x) \in V$ there is $\varepsilon > 0$ with $B_Y(f(x),\varepsilon) \subseteq V$, and continuity at $x$ supplies $\delta > 0$ with $f[B_X(x,\delta)] \subseteq B_Y(f(x),\varepsilon) \subseteq V$, that is $B_X(x,\delta) \subseteq f^{-1}[V]$; as $x$ was arbitrary, $f^{-1}[V]$ is open.

(b) implies (c): let $G \subseteq Y$ be closed; then $Y \setminus G$ is open, so $f^{-1}[Y \setminus G]$ is open by (b), and that set is $X \setminus f^{-1}[G]$, so $f^{-1}[G]$ is closed.

(c) implies (e): let $A \subseteq X$; the set $\overline{f[A]}$ is closed in $Y$, so $G_0 := f^{-1}\big[\overline{f[A]}\big]$ is closed in $X$ by (c), and $A \subseteq G_0$ because $f[A] \subseteq \overline{f[A]}$; hence $\overline{A} \subseteq G_0$ by minimality of the closure, which says exactly $f[\overline{A}] \subseteq \overline{f[A]}$.

(e) implies (a): fix $a \in X$ and a real $\varepsilon > 0$, put $A_\varepsilon := \{x \in X : d_Y(f(x),f(a)) \ge \varepsilon\}$, and suppose no $\delta > 0$ satisfies the continuity condition at $a$ for this $\varepsilon$, that is every ball $B_X(a,\delta)$ contains a point of $A_\varepsilon$; then $a \in \overline{A_\varepsilon}$, so (e) gives $f(a) \in f[\overline{A_\varepsilon}] \subseteq \overline{f[A_\varepsilon]}$, so the ball $B_Y(f(a),\varepsilon)$ meets $f[A_\varepsilon]$ and there is $x \in A_\varepsilon$ with $d_Y(f(x),f(a)) < \varepsilon$, contradicting the definition of $A_\varepsilon$; hence some $\delta > 0$ works, and since $a$ and $\varepsilon$ were arbitrary $f$ is continuous everywhere.

(a) implies (d): let $x_k \to x$ and let a real $\varepsilon > 0$ be given; continuity at $x$ supplies $\delta > 0$ with $f[B_X(x,\delta)] \subseteq B_Y(f(x),\varepsilon)$, and convergence supplies $K$ with $d_X(x_k,x) < \delta$, that is $x_k \in B_X(x,\delta)$, for all $k \ge K$; then $d_Y(f(x_k),f(x)) < \varepsilon$ for all $k \ge K$, so $f(x_k) \to f(x)$.

(d) implies (e): let $A \subseteq X$ and let $y \in f[\overline{A}]$, say $y = f(x)$ with $x \in \overline{A}$; by [L3] there is a sequence $(a_k)$ in $A$ with $a_k \to x$, by (d) $f(a_k) \to f(x)$, and $f(a_k) \in f[A]$ for every $k$, so [L3] applied in $Y$ gives $f(x) \in \overline{f[A]}$.

Steps 1.1, 1.2, 1.3 and 1.4 close the cycle (a), (b), (c), (e), (a), so those four are equivalent; step 1.5 gives (a) implies (d) and step 1.6 gives (d) implies (e), which is one of the four, so (d) is equivalent to them as well; hence all five statements are equivalent.