A proof that $f(K)\le H$ if $K\le G$ for an isomorphism $f: G\to H$. 
Let $f:G \rightarrow H$ be a group isomorphism, and $K \subset G$ be a subgroup. Show that $f(K) \subset H$ is a subgroup.

My attempt:
Let $a \in K$. Then since $K$ is a subgroup, $a^{-1} \in K \rightarrow f(a^{-1}) \in f(K)$. Therefore $f(K)$ is closed under inverse. Similarly, $e \in K$ by definition, so $f(e) \in f(K)$, thus $f(K)$ is closed under identity. Now consider $a,b \in K$. Then $ab \in K$ by definition, and so $f(ab) = f(a)f(b) \in f(K)$ since $f$ is a homomorphism. Therefore $f(K)$ is closed under multiplication and thus a subgroup $\blacksquare$
I do not know if I am playing too fast and loose with applying $f$ in the proof.

A critique of your proof is already given in the comments.

Here's another way to prove the theorem. It uses the one-step subgroup lemma.
We have $e\in K$ since $K\le G$. Thus $f(e)\in f(K)$, so $f(K)$ is nonempty. As the image of an isomorphism, $f(K)$ is a subset of $H$.
Now let $a, b\in f(K)$. Then $a=f(h), b=f(k)$ for some $h, k\in K$. Therefore,
$$\begin{align} 
ab^{-1}&=f(h)f(k)^{-1} \\
&=f(h)f(k^{-1}) \\
&=f(hk^{-1}),
\end{align}$$
but $hk^{-1}\in K$ as $K\le G$. Hence $ab^{-1}\in f(K)$.
Hence $f(K)\le H$.
