I am studying for my qualifiers, and I ran into this question from a previous year's exam.

$\textbf{Question:}$

Consider a two-consumer two-good pure exchange economy. Both preferences are locally non-satiated and convex. Prove or disprove the following statement: if $(x\_1,x\_2)$ and $(\hat{x}\_1,\hat{x}\_2)$ are two different pareto optimal allocations, then the convex combination, $(\alpha x\_1+(1-\alpha)\hat{x}\_1,\alpha x\_2+(1-\alpha)\hat{x}\_2$ MUST also be pareto optimal for any $\alpha\in(0,1)$.

I believe the statement is true, and here is the work for my proof below.

$\textbf{My Proof:}$
By pareto optimality of $x\_i$ and $\hat{x}\_i$: 
$$\not\exists\; x\_i^\star\; s.t.\; u\_i(x\_i^\star)\geq u\_i(x\_i)\;\forall i\; \text{and}\; u\_i(x\_i^\star)> u\_i(x\_i)\;\text{for at least one }i \;\text{or}\;u\_i(x\_i^\star)\geq u\_i(\hat{x}\_i)\;\forall i\; \text{and}\; u\_i(x\_i^\star)> u\_i(\hat{x}\_i)\;\text{for at least one }i$$
$$\implies\;u\_i(\alpha x\_i)\geq u\_i(\alpha x\_i^\star)\;\forall i\;\text{and}\;u\_i((1-\alpha)\hat{x}\_i)\geq u\_i((1-\alpha)x\_i^\star)\;\forall i$$
$$\implies\not\exists\;x\_i^\star\;s.t.\;u\_i(\alpha x\_i^\star+(1-\alpha)x\_i^\star)\geq u\_i(\alpha x\_i+(1-\alpha)\hat{x}\_i)\;\forall i$$
$$\text{and}\;u\_i(\alpha x\_i^\star+(1-\alpha)x\_i^\star)> u\_i(\alpha x\_i+(1-\alpha)\hat{x}\_i)\;\text{for at least one}\;i$$
$$\implies (\alpha x\_i+(1-\alpha)\hat{x}\_i)\;\text{is pareto optimal}$$
$$\blacksquare$$

This proof seemed almost too easy, so I'm wondering if it is correct/rigorous.