Consider a separable, metrizable and locally compact space $\mathrm{X},$ that is not compact. Define $\mathrm{M}_{\mathbf{C}}(\mathrm{X})$ to be the set of complex measures on $\mathrm{X}$ and regard a measure as a linear form (or linear functional) endowed with the vague topology.
To clarify, the vague topology is the topology defined by the seminorms $p_f(\mu) = |\mu(f)|$ where $f$ runs through the set of continuous functions with compact support. Therefore, in what follows we may, and shall, assume that a measure is a linear form defined on continuous functions on $\mathrm{X}$ with compact support.
Consider the Fréchet space $\mathscr{C}_{\mathbf{C}}(\mathrm{X})$ of all $\mathbf{C}$-valued continuous functions on $\mathrm{X}$ endowed with the topology of uniform convergence on compact sets. Define the function $(\mu, g) \mapsto g \cdot \mu$ from $\mathrm{M}_{\mathbf{C}}(\mathrm{X}) \times \mathscr{C}_{\mathbf{C}}(\mathrm{X})$ into $\mathrm{M}_{\mathbf{C}}(\mathrm{X}),$ where $g \cdot \mu$ is the measure define by $f \mapsto \mu(fg).$ Notice that for every continuous $f$ with compact support, $fg$ is continuous with compact support.
One can show, with some work, that if $g_n \to 0$ in $\mathscr{C}_{\mathbf{C}}(\mathrm{X})$ and $\mu_n \to 0$ in $\mathrm{M}_{\mathbf{C}}(\mathrm{X})$ then $g_n \cdot \mu_n \to 0$ in $\mathrm{M}_{\mathbf{C}}(\mathrm{X}).$ In fact, assume every vaguely bounded set has compact and metrizable vague closure (difficult to prove, to be found in every functional analysis book that works weak topologies). Since $\mu_n \to 0$ in $\mathrm{M}_{\mathbf{C}}(\mathrm{X}),$ the set of the $\mu_n$ is vaguely bounded and, therefore, it has a compact and metrizable closure, this implies that $\mu_n$ defines an equicontinuous set of linear forms and so, for every compact subset $\mathrm{K}$ of $\mathrm{X}$ there exists a constant $c_{\mathrm{K}} > 0$ such that $|\mu_n(f)| \leq c_\mathrm{K} \|f\|$ for every continuous $f$ with compact support, from this if $f$ is any continuous function with compact support then $|\mu_n(fg_n)| \leq c_{\mathrm{Supp}(f)} \|fg_n\| \leq c_{\mathrm{Supp}(f)} \|f\| \|g_n\| \to 0.$
I want to show now that $(\mu, g) \mapsto g\cdot \mu$ is not continuous. Having this, it will be a corollary that $\mathrm{M}(\mathrm{X})$ is not metrizable. Any hint? Notice that $g \cdot \mu$ defines a bilinear function on the product of two locally convex spaces, so it is continuous if and only if it is continuous at some pair $(\mu_0, g_0).$
EDIT: I should have said that I meant complex measures. When one deals with positive measures and $\mathrm{X}$ is compact, then the space $\mathrm{M}_+(\mathrm{X})$ of positive measures on $\mathrm{X}$ is separable, metrizable and locally compact. To see, for instance, locally compactness, first remark every continuous function shall have compact support, then consider the constant function $1$ and observe that the neighbourhood of zero in $\mathrm{M}_+(\mathrm{X})$ defined by the relation $|\mu(1)| < 1$ is vaguely bounded; for if $f$ is any continuous function, then positivity of $\mu$ implies $|\mu(f)| \leq \mu(|f|) \leq \mu(1) \|f\| \leq \|f\|,$ and vaguely boundedness follows (recall that a vaguely bounded set here has compact and metrizable vague closure). To see metrizability, consider the fact that $\mathscr{C}_{\mathbf{C}}(\mathrm{X})$ is separable, take a dense sequence $(f_k)$ and define the seminorms $p_k(\mu) = |\mu(f_k)|$ and show that these countably many seminorms define the vague topology, which, by virtue of being separated (Hausdorff), the topology is metrizable. To show separability, one can consider a sequence $(x_n)$ dense in $\mathrm{X}$ and the open balls (for some metric for $\mathrm{X}$) $\mathrm{B}_{n,m} = \mathrm{B}(x_n; \frac{1}{m}),$ then use compactness to extract a subsequence (having the radius fixed) $\mathrm{B} _{{n_j},m}$ ($1 \leq j \leq p_m$), then consider a continuous partition of unity $\varphi_{j,m}$ subordinated to this finite subfamily and consider the set of all $\mathbf{Q}$-linear combinations of Dirac measures at the points $x_n.$ One can clearly write $\mu = \sum \varphi_{j, m} \cdot \mu$ and approximate $\mu(\varphi_{j, m})$ for a rational number $q_{j, m}$ as close as desired.