I don't think your map is injective. Here is an attempt at a recipe for constructing a counterexample.

The ingredients are a $C_1$-field $F$ of characteristic zero and a smooth projective curve $X_0$ over $F$ having non-trivial Brauer group. For a concrete example take $F=\mathbf{C}(t)$ and $X_0$ to be an elliptic curve with non-trivial Brauer group; a calculation of such a curve can be found in O. Wittenberg, *Transcendental Brauer–Manin obstruction on a pencil of elliptic curves*, PDF file here.

**Edit:** What's below is unnecessary. $X_0$ is already a counterexample, since every finite extension of $F$ has trivial Brauer group.

Given these ingredients, let $k$ be the field $F((t))$, let $\mathcal{X}$ be a smooth proper curve over $F[[t]]$ whose special fibre is $X_0$, and let $X$ be the generic fibre of $\mathcal{X}$.

There is a natural injective map $\mathrm{Br}(\mathcal{X}) \to \mathrm{Br}(X)$. Let $\alpha$ lie in the image of this map, and let $P$ be a closed point of $X$. If $R$ is the integral closure of $F[[t]]$ in $k(P)$, then $P$ extends to an $R$-point of $\mathcal{X}$, and $\alpha(P) \in \mathrm{Br}(k)$ lies in the image of $\mathrm{Br}(R)$, which is trivial ($R$ is a Henselian local ring whose residue field is $C_1$). Therefore $\alpha(P)=0$. This holds for all closed points $P$, so $\alpha$ lies in the kernel of your map.

It remains to show that $\mathrm{Br}(\mathcal{X})$ is non-trivial. For any $n$ coprime to the characteristic of $F$, proper base change gives $\mathrm{H}^2(\mathcal{X},\mu_n) \cong \mathrm{H}^2(X_0, \mu_n)$.
Then the Kummer sequence shows that $\mathrm{Br}(\mathcal{X})[n] \to \mathrm{Br}(X_0)[n]$ is surjective. In particular, $\mathrm{Br}(\mathcal{X})$ is non-trivial whenever $\mathrm{Br}(X_0)$ is non-trivial.

*Remark*: if you don't insist that $X$ is a curve, then things are much easier: take a surface over a finite field having non-trivial Brauer group.