[proofplan]
We prove both directions directly from the relative $\varepsilon$-$\delta$ definition of continuity on the subset $E$. If $f$ is discontinuous at $a$, the negation of continuity supplies one fixed tolerance $\varepsilon_0 > 0$ that cannot be met at any scale; choosing the scale $\delta = 1/n$ produces a sequence approaching $a$ whose image values remain at least $\varepsilon_0$ away from $f(a)$. Conversely, we prove the contrapositive: if $f$ is continuous at $a$, then every sequence in $E$ converging to $a$ has image sequence converging to $f(a)$, so no sequential witness for discontinuity can exist.
[/proofplan]
[step:Construct a sequence from the failure of relative continuity]
Assume that $f$ is discontinuous at $a$. By the negation of continuity at $a$ relative to $E$, there exists a number $\varepsilon_0 > 0$ such that for every $\delta > 0$ there exists a point $x \in E$ satisfying
\begin{align*}
|x-a| < \delta
\end{align*}
and
\begin{align*}
|f(x)-f(a)| \ge \varepsilon_0.
\end{align*}
For each $n \in \mathbb N$, apply this assertion with $\delta = 1/n$. This gives a point $x_n \in E$ such that
\begin{align*}
|x_n-a| < \frac{1}{n}
\end{align*}
and
\begin{align*}
|f(x_n)-f(a)| \ge \varepsilon_0.
\end{align*}
Thus $(x_n)_{n=1}^{\infty}$ is a sequence in $E$.
[guided]
Assume that $f$ is discontinuous at $a$. Since continuity is being considered for the map $f: E \to \mathbb R$, the neighbourhood condition is relative to $E$: continuity at $a$ would mean that for every $\varepsilon > 0$ there is a $\delta > 0$ such that every $x \in E$ with $|x-a| < \delta$ satisfies $|f(x)-f(a)| < \varepsilon$.
Negating that statement gives one fixed number $\varepsilon_0 > 0$ with the following property: no matter how small a radius $\delta > 0$ is chosen, there is some point $x \in E$ inside that radius around $a$ for which the desired image estimate fails. In formulas, for every $\delta > 0$ there exists $x \in E$ such that
\begin{align*}
|x-a| < \delta
\end{align*}
and
\begin{align*}
|f(x)-f(a)| \ge \varepsilon_0.
\end{align*}
To turn this pointwise failure at every scale into a sequence, choose the scale $\delta = 1/n$ for each $n \in \mathbb N$. The negated continuity statement then gives a point $x_n \in E$ satisfying
\begin{align*}
|x_n-a| < \frac{1}{n}
\end{align*}
and
\begin{align*}
|f(x_n)-f(a)| \ge \varepsilon_0.
\end{align*}
This defines a sequence $(x_n)_{n=1}^{\infty}$ in the domain $E$.
[/guided]
[/step]
[step:Verify that the constructed sequence converges to $a$ but its image sequence does not converge to $f(a)$]
We first prove $x_n \to a$. Let $\eta > 0$ be given. Choose $N \in \mathbb N$ such that $N > 1/\eta$. If $n \ge N$, then
\begin{align*}
|x_n-a| < \frac{1}{n} \le \frac{1}{N} < \eta.
\end{align*}
Therefore $x_n \to a$.
The sequence $(f(x_n))_{n=1}^{\infty}$ does not converge to $f(a)$, because the number $\varepsilon_0 > 0$ satisfies
\begin{align*}
|f(x_n)-f(a)| \ge \varepsilon_0
\end{align*}
for every $n \in \mathbb N$. Hence the defining convergence condition for $f(x_n) \to f(a)$ fails with tolerance $\varepsilon_0$. This proves the forward implication.
[/step]
[step:Show that continuity sends every convergent domain sequence to a convergent image sequence]
We prove the reverse implication by contraposition. Assume that $f$ is continuous at $a$. Let $(x_n)_{n=1}^{\infty}$ be any sequence in $E$ such that $x_n \to a$. We prove that $f(x_n) \to f(a)$.
Let $\varepsilon > 0$ be given. Since $f$ is continuous at $a$ relative to $E$, there exists $\delta > 0$ such that every $x \in E$ with $|x-a| < \delta$ satisfies
\begin{align*}
|f(x)-f(a)| < \varepsilon.
\end{align*}
Since $x_n \to a$, there exists $N \in \mathbb N$ such that for every $n \ge N$,
\begin{align*}
|x_n-a| < \delta.
\end{align*}
For every $n \ge N$, we have $x_n \in E$, so the continuity estimate applied to $x=x_n$ gives
\begin{align*}
|f(x_n)-f(a)| < \varepsilon.
\end{align*}
Thus $f(x_n) \to f(a)$.
[guided]
We prove the reverse direction in its contrapositive form. Instead of starting from the existence of a bad sequence, assume that $f$ is continuous at $a$ and show that no bad sequence can exist.
Let $(x_n)_{n=1}^{\infty}$ be an arbitrary sequence in $E$ such that $x_n \to a$. To prove $f(x_n) \to f(a)$, we must verify the definition of convergence for the real sequence $(f(x_n))_{n=1}^{\infty}$. So fix an arbitrary tolerance $\varepsilon > 0$.
Continuity of $f: E \to \mathbb R$ at $a$ gives a radius $\delta > 0$ such that every point of the domain $E$ lying within distance $\delta$ of $a$ has its image within distance $\varepsilon$ of $f(a)$. That is, for every $x \in E$, if
\begin{align*}
|x-a| < \delta,
\end{align*}
then
\begin{align*}
|f(x)-f(a)| < \varepsilon.
\end{align*}
Now use the hypothesis $x_n \to a$. Applied with the particular tolerance $\delta > 0$, the definition of sequence convergence gives an index $N \in \mathbb N$ such that for every $n \ge N$,
\begin{align*}
|x_n-a| < \delta.
\end{align*}
Because each $x_n$ belongs to $E$, the continuity implication can be applied to the point $x=x_n$. Therefore, for every $n \ge N$,
\begin{align*}
|f(x_n)-f(a)| < \varepsilon.
\end{align*}
Since $\varepsilon > 0$ was arbitrary, this proves that $f(x_n) \to f(a)$. The argument applies to every sequence in $E$ converging to $a$, so continuity at $a$ rules out the existence of a sequence whose image values fail to converge to $f(a)$.
[/guided]
[/step]
[step:Conclude the equivalence]
The previous step proves that if $f$ is continuous at $a$, then every sequence $(x_n)_{n=1}^{\infty}$ in $E$ with $x_n \to a$ satisfies $f(x_n) \to f(a)$. Taking the contrapositive, if there exists a sequence $(x_n)_{n=1}^{\infty}$ in $E$ such that $x_n \to a$ and $(f(x_n))_{n=1}^{\infty}$ does not converge to $f(a)$, then $f$ is discontinuous at $a$.
Together with the forward implication proved above, this establishes the desired if and only if.
[/step]
