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Martin Leibeck in his textbook A concise introduction to pure mathematics (third edition) brings an example of proof by contradiction in the beginning of the textbook (p.g. 6): (emphasis mine)

Example 1.4

No real number has a square equal to $-1$.

Proof: Suppose the statement is false. This means there is a real number, say $x$, such that $x^2=-1$. However it is a general fact about real numbers that a square of any real number is greater than or equal to $0$. Hence $x^2 \ge 0,$ which implies that $-1 \ge 0$. This is a contradiction.

My question:

Does the Italic section of the proof have valid steps necessary for proving such a statement? Is saying, "However it is a general fact about..." a valid step in writing a proof? Wouldn't it be necessary to define what a "real number" is?

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    I praise you for being rigorous and seeing the apparent gap in the proof. However, these gaps won't be filled until you're a little older in mathematics, so bear with the toothache for some time, your teeth will be filled once you have fulfilled the mathematical prerequisites. The definition of a real number is just going through a real dentist chair, so it would really slow you down at this stage. Continue reading on, and look up a bunch of references (for real numbers, Rudin would do) to fill gaps. – 2017-01-24
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    I am not sure if this book introduces closure properties. The operator $\times$ is closed for $\mathbb{R}$ and therefore $x \times x$ which is the square operator is also closed in $\mathbb{R}$. Three cases can be considered when $x<0$ $x=0$ and $x>0$. $x>0$ and $x=0$ are trivial cases, for $x<0 $ however $x \times x$ would be $>0$. I would however not call this a general fact in a proof. – 2017-01-24
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    Statements like "it's a general fact" or "it is well-known" may depend on your audience. In terms of a correct logic, they also depend on what can be presumed to have been previously proven. The axioms for an ordered field imply that $x^2=(-x)^2\geq 0 >-1$.... I think the author presumes that to be already done, and is illustrating the method of proof by contradiction : That the existence of a real $x$ such that $x^2+1=0$ implies a paradox. – 2017-01-24
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    If that proof is actually to rigorously prove a pertinent statement, I'd say you are absolutely correct. And I'd say that proof is meaningless as it is essentially claiming what it is attempting to prove. But if it attempting to illustrate the principals of proof by contradiction, by showing a trivial and obvious example, I can't fault it too heavily. – 2017-01-24

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Doing so would be extremely tedious and greatly slow down the process of learning new things. Most math books tell you who the intended reader is, this is so that you have the appropriate prerequisites to be able to fill gaps in proofs.

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"General" seems to mean "well-known," and yes, with that meaning the statement is valid in a proof provided the fact is indeed well-known. When learning, you can so describe stuff that appears in textbooks for prerequisite courses. When it's the kind of stuff that appears in the textbook for the course you're currently in, it's better to give a cite to a place in the textbook or a theorem name (e.g., Lebesgue Dominated Convergence).

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No. And in this case it's especially egregious as this prove seems to boil down to: proof that $x^2\ne -1:$ c'mon, you know it is positive!

BUT note this is an example. Not actually a theorem. The purpose of this is not to show a rigorous proof, but simply to introduce the student to the concept of a proof by contradiction. Rather than proving something significant, they showed an example where the student really already knew why. This is so it is the concept of contradiction that is being shown. Not the rigor.

I often use silly examples. Such as Proof that I do not own a unicorn: assume I own a unicorn. Everyone who owns a unicorn is extremely happy. Therefore I am happy. But look at me; I'm obviously miserable. So it'd be impossible for me to own a unicorn.

That's not really a valid argument. All the statements are presumed as given and not verified.

But it shows the idea that we presume the opposite, we deduce, we come to a contradiction. So the opposite is impossible.

That's all that is relevent.

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No, it is not. Nothing has been proven in the proof you posted. The general fact should be replaced with something like:

Every square of a real number is positive, because

  • if $x >0$ then $x \cdot x > 0$
  • if $x <0$ then $x \cdot x = (-y)(-y) >0$ for $ x \space \epsilon \space \mathbb{R}$ and $y = -x$