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I only know that they are $T_2$ and second countable.

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    Yes. I don't have time to write full answer, but consider the projections $S^n\xrightarrow{\pi} \mathbb{R}P^n$ and $S^{2n+1}\xrightarrow{\pi}\mathbb{C}P^n$. Picking $p,q$ in either $\mathbb{R}P^n$ or $\mathbb{C}P^n$, define $d(p,q) = \overline{d}(\pi^{-1}(p), \pi^{-1}(q))$ where $\overline{d}$ is the usual metric on $S^n$ and $\pi^{-1}(p)$ denote the (compact) preimage set of $p$: all the points in $S^n$ which map down to $p$.. – 2017-02-27
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    I failed to find out if your spaces are also normal ($T_4$)... if they are and you can also show that they are Housdorff, you can apply Uryssohn's metrization theorem. Since there is a homeomorphism between $RP^n$ and a n-dimentional sphere (with a twist), both statements might actually be true ($T_4$ and Housdorff) – 2017-02-27
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    Yes, but I also failed to show that they are at least $T_3$. If they are then one can also use Urysohns theorem – 2017-02-27
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    Don't you know they are compact? Indeed, they may be seen as particular quotients of a (real or complex) $n$-dimensional sphere. Now, compact and second countable together imply metrizable. – 2017-02-27
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    Really? I did know that they are compact, but I did not know that compact + second countable implies metrizability – 2017-02-27
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    @AndreiKh compact plus Hausdorff gives normal hence regular. Now Urysohn's metrization theorem does the rest. You do need the Hausdorff condition (as a countable co-finite space shows, which is compact $T_1$ and second countable, but not metrisable) – 2017-02-27

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Since $\mathbb{RP}^n$ and $\mathbb{CP}^n$ are compact and $T_2$, they are regular ($T_1 + T_3$). Hence, by Urysohns metrization theorem, they are also metrizable.