Prove that $\sqrt{5}$ exists Prove that $\sqrt{5}$ exists; in other words prove that there exists a positive number $x\in \mathbb R$ satisfying $x^2=5$

Here's what I've done:

I let $A= \{x&gt;0:x^2\leq 5\}$

We know that $A$ is not empty because clearly $2$ is in it: $2^2&lt;5$ and we also know that $A$ is bounded by $3$ because

$$x&gt;3\implies x^2&gt;9$$ so by completeness, $\sup A$ exists. I let $α = \sup A$

So now I'm trying to prove that $α^2=5$ (that there exists a positive number $α\in \mathbb R$ satisfying $α^2=5$) and to do this I'm going to try to show that $α^2&lt;5$ and $α^2&gt;5$ are impossible.

Case 1: $α^2&lt;5$ to do this I'm assuming a proof by contradiction will work.

So, suppose $α^2&lt;5$ then ...

Here's where I'm a bit lost, I don't know how to proceed with a proof by contradiction here. If anyone can give me hints or explain what I should do next that would be really appreciated.
If we use your strategy, we need to show that we cannot have $\alpha^2\lt 5$, and we cannot have $\alpha^2\gt 5$. We do this with minimal machinery. That makes the argument much longer than if there is some already developed theory to appeal to. 

Suppose that $\alpha^2\lt 5$. Let $\alpha^2+\delta=5$ for some positive $\delta$. Note that $\delta \le 1$ and $\alpha\ge 2$. Then 
$$\left(\alpha +\frac{\delta}{4\alpha}\right)^2=\alpha^2+\frac{\delta}{2}+\frac{\delta^2}{16\alpha^2}=\alpha^2+\delta\left(\frac{1}{2}+\frac{1}{64}\right)\lt 5.$$
It follows that $\alpha +\frac{\delta}{4\alpha}\in A$, contradicting the fact that $\alpha$ is an upper bound for $A$. 

Suppose now that $\alpha^2\gt 5$.  Let $\alpha^2=5+\delta$ for some positive $\delta$. We leave it to you to find a positive $x$ smaller than $\alpha$ such that $x^2\gt 5$, contradicting the fact that $\alpha$ is the least upper bound of $A$. Hint: Something like what we did will work, except that we must subtract some small number from $\alpha$, instead of adding a small number, as we did earlier.
Hint
Show that $f(x) := x^2-5$ is continuous on $\mathbb R$ and use the intermediate value thm. on
$(2, -1) \in {\rm graph}(f)$ and $(3,4) \in {\rm graph}(f)$