The general idea of the proof is right, and is basically how we prove compactness once we have completeness and soundness. Let's first make a few definitions and state some results to be clear what we are talking about.
As usual, for a theory $\Gamma$, the notation $\Gamma \vdash \varphi$ means there exists a formal derivation of $\varphi$ from $\Gamma$ (with respect to some classical proof system, it does not really matter which one). The notation $\Gamma \models \varphi$ means that $\varphi$ is valid in all models of $\Gamma$.
Definition. A theory $\Gamma$ is formally consistent if $\Gamma \not \vdash \bot$.
So $\Gamma$ is formally inconsistent if $\Gamma \vdash \bot$, which is equivalent to $\Gamma \vdash \varphi$ and $\Gamma \vdash \neg \varphi$ for some formula $\varphi$ (this is just to link this definition to what is used in the question).
Henkin's lemma. If $\Gamma$ is formally consistent, then it has a model.
Soundness theorem. If $\Gamma \vdash \varphi$, then $\Gamma \models \varphi$.
Completeness theorem. If $\Gamma \models \varphi$, then $\Gamma \vdash \varphi$.
Proof. We can actually easily prove the completeness theorem from Henkin's lemma (fair note: Henkin's lemma is far from trivial to prove). We will prove the contraposition, so suppose $\Gamma \not \vdash \varphi$. Then $\Gamma \cup \{\neg \varphi\}$ is formally consistent. By Henkin's lemma there is then a model of $\Gamma \cup \{\neg \varphi\}$. This is then in particular a model of $\Gamma$ where $\varphi$ is not valid, so $\Gamma \not \models \varphi$, as required.
Now we can prove compactness from soundness and completeness. Personally I like this proof because it gives a great intuition as to why the compactness theorem is true. Once we accept that semantics (i.e. "$\models$") and provability (i.e. "$\vdash$") coincide, then we can say that any contradiction must be derivable. Since derivations are finite it must be derivable from a finite set of assumptions. Let's make this precise.
Compactness theorem. If every finite subset of a theory $\Gamma$ has a model, then $\Gamma$ has a model.
Proof. Suppose not. So $\Gamma$ has no model, but every finite subset does have a model. Since $\Gamma$ has no model, we have (vacuously) $\Gamma \models \bot$. So by completeness $\Gamma \vdash \bot$ (alternatively: by the contraposition of Henkin's lemma). Since (formal) derivations are finite, there must be a finite subset $\Delta \subseteq \Gamma$ such that $\Delta \vdash \bot$. By soundness $\Delta \models \bot$, and as no model can satisfy $\bot$ we see that $\Delta$ has no models. This contradicts our assumption that every finite subset of $\Gamma$ has a model, so we conclude that $\Gamma$ must have a model.
Best Answer
You did use the contrapositive here, but of the compactness theorem itself. Your proof is fine like this (although the last bit of the last sentence is missing: "... such that $\Delta \cup \{\neg \phi\}$ is inconsistent").
The other approach would be to prove the contrapositive of the statement you wish to prove. That is: if for no finite $\Delta \subseteq \Gamma$ we have $\Delta \models \phi$, then $\Gamma \not \models \phi$. To do this we claim that $\Gamma \cup \{ \neg \phi \}$ is consistent, which we prove by using the compactness theorem. Let $\Delta \subseteq \Gamma$ be finite, then by our assumption $\Delta \not \models \phi$, so $\Delta \cup \{\neg \phi\}$ is consistent. Hence every finite subset of $\Gamma \cup \{ \neg \phi \}$ is consistent and we are done.