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Let us suppose that $\lambda \in Card$ where $Card$ is the class of all infinite cardinals.

If $\kappa>\lambda$ and $cof(\kappa)>\lambda$, why does the last inequality imply that every function from $\lambda$ into $\kappa$ is bounded by some ordinal $v<\kappa$?

From my point of view, if we consider any such function, shouldn't we get $f(x)<\delta_x$ where $\delta_x$ is contained in the cofinal subset of $\kappa$ having minimal cardinality (from the way how we defined cofinality) and for any $x\in\lambda$ i.e. $x<\lambda$. Hence taking maximum of all such $\delta_x$ would consequently result the same conclusion as above.

Now I wonder where did I get wrong.

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    If $f\colon \lambda\to \kappa$ is unbounded, then the function $f$ is cofinal, hence cof$(\kappa)\leq \lambda$. Right? – 2017-02-22
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    @RenanManeliMezabarba You mean,in that case we would have that $f[\lambda]$ is cofinal in $\kappa$, hence cof$(\kappa)\leq$card$f[\lambda]\leq\lambda$? – 2017-02-22
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    How does your argument fail in the case that the cofinality of $\kappa$ does _not_ exceed $\lambda$? How do you know that the maximum of a subset of a cardinal is contained in that cardinal? – 2017-02-22
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    Since $\kappa\geq\operatorname{cf}(\kappa)$, it is enough to require that $\operatorname{cf}(\kappa)>\lambda$. As to how and why, try working out the specifics when $\kappa=\omega_1$ and $\lambda=\omega$. This should give you an intuition as to why this theorem is true. – 2017-02-22
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    Note also that $\mathrm{cof}(\kappa) \leq \kappa$, so the inequality $\kappa > \lambda$ is unnecessary given that $\mathrm{cof}(\kappa) > \lambda$. – 2017-02-25
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    @RenanManeliMezabarba is exactly right. – 2017-02-25

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