I have a difficulty trying to prove the following proposition. Any help would be greatly appreciated.
$\textbf{Prop.}$ Let $(X,d_1)$ and $(Y,d_2)$ be two metric spaces. A function $f:X\rightarrow Y$ is continuous at $x_0\in X$ if and only if every sequence $(x_n)_{n\in\mathbb{N}}$ in $X$ that converges to $x_0$ has a subsequence $(x_{n_k})_{k\in\mathbb{N}}$ such that $f(x_{n_k})\rightarrow f(x_0)$.
The forwards direction is obvious using the usual sequential characterization of continuity.
For the reverse implication I tried arguing as follows: suppose, for contradiction, that $f$ is not continuous at $x_0$. Then there is some sequence $(x_n)_{n\in\mathbb{N}}$ in $X$ that converges to $x_0$ but $(f(x_n))_{n\in\mathbb{N}}$ does not converge to $f(x_0)$. I'm stuck here since I think $(f(x_n))_{n\in\mathbb{N}}$ may still have a subsequence that converges to $f(x_0)$, right?