Suppose that $(X,d_{X})$ and $(Y,d_{Y})$ are metric spaces. Let $f:X\to Y$ be a function, and let $x_{0}\in X$ be a point of $X$. Then the following two statements are logically equivalent:
(a) $f$ is continuous at $x_{0}$.
(b) Whenever $x^{(n)}$ is a sequence in $X$ which converges to $x_{0}$ with respect to the metric $d_{X}$, the sequence $f(x^{(n)})$ converges to $f(x_{0})$ with respect to the metric $d_{Y}$.
MY ATTEMPT (EDIT)
Let us prove the implication $(a)\Rightarrow(b)$ first.
According to the definition of limit, for every $\varepsilon > 0$, there is a real number $\delta > 0$ such that
\begin{align*}
d_{X}(x,x_{0}) < \delta \Rightarrow d_{Y}(f(x),f(x_{0})) < \varepsilon
\end{align*}
On the other hand, if $x^{(n)}$ converges to $x_{0}$, then for every $\delta > 0$ there is a natural number $N\geq 0$ such that
\begin{align*}
n\geq N\Rightarrow d_{X}(x^{(n)},x_{0}) < \delta
\end{align*}
Let $\varepsilon > 0$ and $x = x^{(n)}$. Then there is a $\delta > 0$ and a natural number $N\geq 0$ such that
\begin{align*}
n\geq N \Rightarrow d_{X}(x^{(n)},x_{0}) < \delta \Rightarrow d_{Y}(f(x^{(n)}),x_{0}) < \varepsilon
\end{align*}
whence we conclude that $f(x^{(n)})\to f(x_{0})$.
We may now prove that $(b)\Rightarrow(a)$.
Let us prove it by contrapositive. Then there is a $\varepsilon > 0$ such that for every $\delta > 0$ there exists and $x$ such that $d_{X}(x,x_{0}) < \delta$ and $d_{Y}(f(x),f(x_{0})) \geq \varepsilon$.
If we take $\delta = 1/n$, there corresponds a $x = x^{(n)}$, whence we conclude by the squeeze theorem that
\begin{align*}
0\leq d_{X}(x^{(n)},x_{0})\leq 1/n \Rightarrow d_{X}(x^{(n)},x_{0})\rightarrow 0
\end{align*}
But we do also have that
\begin{align*}
d_{Y}(f(x^{(n)},f(x_{0})) \geq \varepsilon
\end{align*}
which ensures that $f(x^{(n)})$ does not converge to $f(x_{0})$. Then the proposed result holds.
EDIT
The definition of continuity which I've been given is:
$f:(X,d_{X})\to(Y,d_{Y})$ is continuous at $x_{0}$ iff for every $\varepsilon > 0$, there is a $\delta > 0$ such that
\begin{align*}
d_{X}(x,x_{0}) < \delta \Rightarrow d_{Y}(f(x),f(x_{0})) < \varepsilon
\end{align*}
Best Answer
Two small comments for the first part. First, what is $m$? Secondly, it is more clear to fix the quantities. When you write a "for all $\epsilon > 0$" statement, the $\epsilon$ only makes sense in the context of this statement, and it is lost afterwards. So it would be nicer to say "Let $\epsilon > 0$. There exists $\delta$... There exists $N$... So for $n \geq N$..."
Mostly, for the second part, there are problems with how you write the negation of a statement. The fact that $f$ is not continuous at $x_0$ is not what you claim: it is that there is an $\epsilon > 0$ such that for any $\delta > 0$, there exists $x$ such that $d_X(x,x_0) < \delta$ and $d_Y(f(x),f(x_0)) > \epsilon$.
Then, a more straightforward way to show that (b) $\implies$ (a) is by contrapositive: in other words, assume that $f$ is not continuous at $x_0$, and show that there is a sequence $(x^{(n)})$ such that $x^{(n)} \to x_0$ but $f(x^{(n)})$ does not converge to $f(x_0)$. For this, take $\epsilon > 0$ as above, and $\delta = 1/n$. Can you continue?