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I feel this is an important question for me to understand to get the whole hang of it.

Everyone except one student in your class has an internet connection

Given I(x) = x has an internet connection, domain: set of all students in my class.
My solution is:
∀x ∃y {(x ≠ y)⟶I(x) ⋀ ㄱI(y) ⋀ ∀k(ㄱI(k)⟶(y=k))}

Thought pattern:
For every student x,there is a student y for which if x is not y, then x has internet connection. The y guy(s) doesn't have internet connection, and to make y unique, add the later stuff ...⋀ ∀k(ㄱI(k)⟶(y=k))

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    You should have $\exists y$ first, because the student that doesn't have internet is fixed and doesn't depend on what student $x$ we're checking. – 2017-02-19
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    The doubt is how to make that y unique. Exactly one student NOT having internet connection. Did I do it right there? – 2017-02-19
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    @Arthur yeah, you are right. This will be read as 'for every student x, there is a y' which doesn't make sense. – 2017-02-19

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Your expression doesn't tell us anything. Note that any formula that reads "$\forall x\exists y((x\neq y)\to\cdots)$" is necessarily true because $y=x$ is a $y$ that exists and makes the contents of the parenthesis true.

Rather, I would think like this: There is a student that does not have internet, and every other student has internet. It would read like this: $$ \exists y(\lnot I(y)\land\forall x(x\neq y\to I(x))) $$

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    But I need EXACTLY ONE student not having internet! – 2017-02-19
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    Or can I simply give ∃!y (¬I(y))? hehe – 2017-02-19
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    No, this makes it exactly one, because $\forall x(x\neq y \to I(x))$ says that every other student has internet. – 2017-02-19
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    Got it. Thanks. – 2017-02-19