Our class's lecture notes approached it indirectly so I want my worked to be checked here.
Proposition: Let $f:(X,\rho)\rightarrow (Y,\sigma)$ be a continuous function between two metric spaces. Then $f$ is continuous at $x\in X$ if and only if for every sequence $(x_n)$ that converges to $x$, then the corresponding sequence $(f(x_n))$ converges to $f(x)$.
Proof attempt:
Suppose $\sigma(f(x_n),f(x))<\epsilon$. Since $f$ is continuous at $x\in X$ then $\exists \delta >0$ such that for all $\alpha\in X$, we have that if $\rho(x,\alpha)<\delta$, then $\sigma(f(x),f(\alpha))<\epsilon$. But since $(x_n)$ is convergent to $x$, then there exists $N\in\Bbb{N}$ such that $\forall n\geq N$, $\rho(x,x_n)<\delta$. Thus, for all $n\geq N$, $\sigma(f(x_n),f(x))<\epsilon$. Hence, $(f(x_n))$ converges to $f(x)$. Conversely, if $(f(x_n))$ converges to $f(x)$ then $\forall\epsilon'>0\exists M\in\Bbb{N}$ such that $\forall m\geq M$, $\sigma(f(x_m),f(x))<\epsilon'$. Note that $\forall m\geq M$, $\rho(x,x_m)<\delta'$, for some $\delta'>0$. Since $(x_n)$ is convergent to $x$, $\exists M_1\in\Bbb{N}$ so that for all $i\geq M_1$, $\rho(x,x_i)<\delta'$. Therefore, $f$ is continuous at $x$.
The converse (if the sequence condition holds at $x$, then $f$ is continuous at $x$) has to be an indirect proof. The forward direction you almost got right, except that the first sentence is nonsense: you assume $x_n \to x$ and you know continuity $\epsilon-\delta$ definition) and you need to show that $f(x_n) \to f(x)$, so to show the last with the definition of convergence in $Y$ you have to start by picking $\epsilon>0$.
By continuity of $f$ at $x$, there exists a $\delta>0$ such that for all $x' \in X$: if $d(x,x') < \delta$ we know that $d(f(x'), f(x)) < \epsilon$. So in order to get the $f(x_n)$ $\epsilon$-close to $f(x)$ we know that $x_n$ has to be $\delta$-close to $x$, and this will happen eventually: there exists $N \in \mathbb{N}$ such that for all $n \ge N$ we have that $d(x_n, x) < \delta$. But then for this same $N$ we thus have that $d(f(x_n), f(x)) < \epsilon$ for all $n \ge N$, as required. You can see how the definition of convergence nicely interacts with the continuity: both are about eventual approximation.
For the reverse we have to show that for each $\varepsilon > 0$ there exists some $\delta >0$ such that for all $x'$ $\delta$-close to $x$, $f(x')$ is $\varepsilon$-close to $f(x)$. And all are given is a statement about convergent sequences. Now going for a contradiction we at least get a positive statement we can work with: if $f$ is not continuous at $x$, there exists some fixed $\epsilon >0$ such that no $\delta>0$ will "work", so for all $\delta>0$, there is some $x(\delta)$ (the point depends on how we choose the $\delta$, hence the notation) such that $d(x(\delta), x) < \delta$ but still $d(f(x(\delta)), f(x)) \ge \epsilon$.
Now a standard way to make a sequence in $X$ that converges to $x$, is to pick a point $x_n$ with $d(x_n, x)< \frac1n$ (or some other term in $n$ that goes to $0$ for larger and larger $n$). We can do this by our contradiction assumption: let $x_n = x(\frac1n)$ in the above. Then we know that $x_n \to x$ so now the sequence assumption can finally be used! We conclude that $f(x_n) \to f(x)$ but his is a contardiction, as in that case $f(x_n)$ should get $\epsilon$-close to $f(x)$ for large enough $n$, while for all $n$ these are at least $\epsilon$ apart, contradiction. So $f$ is continuous at $x$.
So you should have to product a $\delta>0$ for this proof of continuity. You cannot do that easily, but if you assume no such $\delta$ exists, we can construct a sequence on which to apply the sequence assumption. The indirectness seems quite unavoidable to me.