Here is what I think is the simplest strategy. I'm only giving a  sequence of lemma which lead to the result and I think they are all easy enough, but maybe a little teddious to write down (but let me know if something is not as easy as I thought! note that extensivity is used in many places)

**Lemma 1:** IF $A \subset X$ is a subobject and $R \subset A \times A$ is an equivalence relation on $A$, then $R \cup \Delta_X$ is an equivalence relation on $X$.

**Lemma 2:** In the setting of the previous lemma $X \to X/(R \cup \Delta_X)$ is the pushout of $A \to A/R$.

**Lemma 3:** For any object $U \coprod U \to U$ is the quotient of $U \coprod U$ by the relation 

$$U \coprod U \coprod U \coprod U \to (U \coprod U)^2 = U^2 \coprod U^2 \coprod U^2 \coprod U^2$$

**Lemma 4:** For any monomorphism $U \to X$,the pushout $X \coprod_U X$ is the quotient of $X \coprod X$ by an equivalence relation that can be written as

$$ X \coprod U \coprod X \coprod U \to (X \coprod X)^2 = X^2 \coprod X^2 \coprod X^2 \coprod X^2 $$

*Hint:* use that $X \coprod X \to X \coprod_U X $ is a pushout of $U \coprod U \to U$ and all the lemmas above.


**Lemma 5:** $U$ is the equalizer of the two maps $X \rightrightarrows X \coprod_U X$.

*Hint:* Lemma 4 and effectiveness of equivalence relations lets you compute the pullback of $X \coprod X \to X \coprod_U X$ with itself. From there you can deduce this other limit using formal manipulation of limits.


**Conclusion:** We have shown that an arbitrary monomorphism can be written as an equalizer. In particular it is a regular monomorphisms. This implies that the category is balanced.


**Why I think (but could be wrong) there isn't a simplest proof:**
If I'm not mistaken, an exact category need not be balanced. For example the category of commutative rings is exact but not balanced, so the sort of argument you were doing using only quotient and pullback will not work. You really need to also use coproducts and extensivity in the proof, and I don't see another way to do  this.


**What is probably a simpler way to do this** I assumed you wanted a direct diagramatic proof. But at this point I feel like at this point it is simpler to prove some form of Giraud's theorem and use it. Even without assuming local presentability, by (the proof of) Giraud's theorem any finite diagram in your category can be lifted to some Grothendieck topos T with a functor $T \to \mathcal{E}$ preserving colimits and finite limits. One can then just proves the theorem in a Grothendieck topos, i.e. a category of sheaves, where everything becomes a lot simpler, even doing the argument above, and you can go even quicker by showing there is a subobject classifier.