Pseudo Proofs that are intuitively reasonable
Cayley-Hamilton Theorem. Let $A$ be an $n\times n$ matrix, and let $f(t)$ be its characteristic polynomial. Then $f(A)=0$.
"Proof." $f(t) = \det(A-tI)$. Therefore, $f(A) = \det(A-AI) = \det(A-A) = \det(0) = 0$.
Issues. One problem may not be obvious... the equation "$f(A)=0$" is really saying that the matrix we get via the evaluation (by identifying the underlying field with the subring of scalar matrices) is the zero matrix. However, the "proof" claims to prove that $f(A)$, which is supposed to be a matrix, is equal to the value of a determinant, which is a scalar.
As to why it "gives the right answers"... well, because the theorem is true. I am reminded of what Hendrik Lenstra once said in class after presenting an idea for a proof and explaining why it didn't quite work:
The problem with incorrect proofs to correct statements is that it is hard to come up with a counterexample.
$$\begin{align*} x^{x^{x^{\scriptstyle\ldots}}} &= 2\\ x^{\left(x^{x^{x^{\scriptstyle\ldots}}}\right)} &= 2\\ x^2 &= 2\\ x &= \sqrt{2}\\ \end{align*}$$
This is one that comes up on math.SE from time to time.
Proposition: $[0, 1]$ has the same cardinality as $[0, 1]^2$.
"Proof 1." Write $(x, y) \in [0, 1]^2$ in their binary expansions $x = 0.x_1 x_2 x_3 ..., y = 0.y_1 y_2 y_3 ...$, and consider the map which sends $(x, y)$ to $0.x_1 y_1 x_2 y_2 x_3 y_3 ...$.
Of course the problem with this proof is that the map is not well-defined since numbers do not have unique binary expansions, e.g. we have $0.1 = 0.0111...$.
"Proof 2." Repeat Proof 1, but this time don't allow any terminating binary expansions except $0.0$; instead, when a number has two binary expansions, pick the one which ends with infinitely many ones.
But now our map isn't surjective! The restriction on binary expansions above means, for example, that we can never hit $0.011010101...$.
A neat fix is just to show that $[0, 1]$ has the same cardinality as $\{ 0, 1 \}^{\mathbb{N}}$, which is straightforward to do by Cantor-Bernstein-Schroeder, and then the above argument actually works.