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This proof should be easy but I get stuck because I don't know how to deal with the infimum in this case.

I want to prove that $$d(x,A)\le d(x,y)+d(y,A) $$ with $$d(x,A)=\text{inf}(\{d(x,a):a\in A\})$$

My attempt: Let $y\in X$. We have $d(x,a)\le d(x,y)+d(y,a) $ for any $a\in A$.

How can I get the inequality for the infimum from this?

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    Show that for every $\varepsilon > 0$ you have $d(x,A) \leqslant d(x,y) + \bigl(d(y,A) + \varepsilon\bigr)$. – 2017-01-18

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For all $a\in A$ we have by triangle inequality

$$d(x,A)\le d(x,a)\le d(x,y)+d(y,a)$$ so $$d(x,A)-d(x,y)\le d(y,a),\;\forall a\in A$$ hence $$d(x,A)-d(x,y)\le d(y,A)$$

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    You seem to have used $d(y,a)\le d(y,A)$... – 2017-01-18
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    Otherwise, how can you justify the last step? Just because $d(x,A)-d(x,y)\le d(y,a),\;\forall a\in A$ does that mean that $d(x,A)-d(x,y)\le d(y,A)$? There may not exist an $a\in A$ such that $d(y,A)=d(y,a)$... – 2017-01-18
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    @XicoSimThe answer is correct. Look at the definition of the infimum. – 2017-01-18
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    @juniven That's what I've done, and $d(y,A)\le d(y,a)$, not the other way around. – 2017-01-18
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    I had this problem with proofs like these numerous times: one can have $d(y,A)< d(y,a) \forall a\in A$, so the last step does not seem legit to me. – 2017-01-18
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    @XicoSim If you have $C \leqslant d(y,a)$ for all $a\in A$, then you also have $C \leqslant \inf\limits_{a\in A} d(y,a)$. – 2017-01-18
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    @DanielFischer Ok, That does not seem obvious to me, though, because of what I said on my previous comment: the infimum of a set can be smaller than all the elements of the set, so guaranteeing $C\le b \,\,\forall b \in B$ does not seem to guarantee that $ C\le \text{inf}B$ – 2017-01-18
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    @XicoSim Okay, $d(x,A)-d(x,y)$ serves as a lower bound of the set $\{d(y,a):a\in A\}$. So , by the definition of $d(y,A)$, the last line holds. – 2017-01-18
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    @XicoSim If $\inf B < D$, then in particular there is a $b\in B$ with $b < D$. So if $C \leqslant b$ for all $b\in B$, then $C \leqslant \inf B$. – 2017-01-18
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Let $\epsilon>0$. Then $d(y,A)+\epsilon$ can not be a lower bound of the set $\{d(y,a):a\in A\}$. Thus, there exists $w\in A$ such that

$$d(y,w)

Thus, $$ \begin{align} d(x,A)&\leq d(x,w)\\ &\leq d(x,y)+d(y,w)\\ &0.$$ Thus, $$d(x,A)\leq d(x,y)+d(y,A).$$