Let $f:X\rightarrow Y$ be a continuous map of metric spaces. Show that if $A\subseteq X$ is compact, then $f(A)\subseteq Y$ is compact.
I am using this theorem: If $A\subseteq X$ is sequentially compact, it is compact. Also this definition: A set $A\subseteq X$ is sequentially compact if every sequence in $A$ has a convergent subsequence in $A$.
Attempt at a proof:
Let $\{y_n\}\subseteq f(A)$. Since $f$ is continuous, $\{y_n\}=f(x_n)$ for some $\{x_n\}\subseteq A$. If $A\subseteq X$ is compact, every sequence $\{x_n\}\subseteq A$ has a subsequence that converges to a point in $A$, say $\{x_{n_k}\}\rightarrow a\in A$. Since $f$ is continuous, $f(x_{n_k})\rightarrow f(a)\in f(A)$. Then $f(x_{n_k})\subseteq f(A)$ is a convergent sequence in $f(A)\implies f(A)$ is compact since $\{y_n\}\subseteq f(A)$ was arbitrary.