FALSE: every real number has a real square root

False claim: every real number has a real square root, that is, for every $x \in \mathbb{R}$ there exists $y \in \mathbb{R}$ with $y^{2} = x$.

The true statement is Square roots exist: a unique $\sqrt{a} \ge 0$ with $(\sqrt{a})^2 = a$; the positives are $\{x^2 : x \neq 0\}$, which produces a square root only for $x \ge 0$, and its generalisation Existence and uniqueness of $n$-th roots: a unique $a^{1/n} \ge 0$ with $(a^{1/n})^n = a$, which produces an $n$-th root only for $x \ge 0$. The nonnegativity hypothesis in both is load bearing, not decoration.

• What fails is evenness of the exponent, not the taking of roots. Odd roots of negative numbers do exist in $\mathbb{R}$. The library has no general theory of parity, so fix the local abbreviation: call a natural $n$ odd when $n = 2k + 1$ for some natural $k$. Every odd $n$ satisfies $n \ge 1$.
• First, $(-1)^{n} = -1$ for odd $n$, which is a computation and not an unstated induction: $(-1)^{2k+1} = \big((-1)^{2}\big)^{k} \cdot (-1)$ by the addition and iterated-power laws for natural exponents (Laws of integer exponents, claim 1), $(-1)^{2} = (-1)(-1) = 1$ ($(-1)(-1) = 1$), and $1^{k} = 1$ (Monotonicity of $x \mapsto x^n$ and of $n \mapsto a^n$, claim 4); so the product is $1 \cdot (-1) = -1$. Consequently $(-u)^{n} = (-1)^{n} u^{n} = -\,u^{n}$ for every $u \in \mathbb{R}$ and odd $n$ (Laws of integer exponents, Sign rules for products: $(-a)b = -(ab)$ and $(-a)(-b) = ab$).
• Every real is an $n$-th power, for odd $n$. For $a \ge 0$ take $x = a^{1/n}$ ([A3], Existence and uniqueness of $n$-th roots: a unique $a^{1/n} \ge 0$ with $(a^{1/n})^n = a$). For $a < 0$ take $x := -\,(-a)^{1/n}$, which is legitimate because $-a > 0$ (Sign rules for products and monotonicity of multiplication), and then $x^{n} = -\big((-a)^{1/n}\big)^{n} = -(-a) = a$ by the previous item.
• And the $n$-th power map is injective for odd $n$, which the strict increase on $\{x \ge 0\}$ alone does not give, since a sign statement is not a monotonicity statement. The map is strictly increasing on the whole line. On $\{x \ge 0\}$ that is Monotonicity of $x \mapsto x^n$ and of $n \mapsto a^n$, claim 2. If $x < y \le 0$ then $0 \le -y < -x$, so $(-y)^{n} < (-x)^{n}$ by that same claim, and negating gives $x^{n} = -(-x)^{n} < -(-y)^{n} = y^{n}$ (Sign rules for products and monotonicity of multiplication). If $x < 0 \le y$ then $-x > 0$ gives $(-x)^{n} > 0$, so $x^{n} = -(-x)^{n} < 0$, while $y^{n} \ge 0$ (Monotonicity of $x \mapsto x^n$ and of $n \mapsto a^n$, claim 1). So $x < y$ always implies $x^{n} < y^{n}$; the map is injective, and with the surjectivity above it is a bijection of $\mathbb{R}$ onto itself for every odd $n$.
• None of this rescues the even case, and that is the point of the item: for even $n$, meaning $n = 2k$, the same computation gives $(-1)^{n} = \big((-1)^{2}\big)^{k} = 1^{k} = 1$, so $x^{n} = (-x)^{n} \ge 0$ for every $x$, powers of both signs land in $\{x \ge 0\}$, and the refutation above applies verbatim with $n = 2$.