Let $B(H)$ be the set of bounded operators on a Hilbert space $H$.
I know that $u_{\alpha}\longrightarrow u$ in S.O.T if and only if $u_{\alpha}(x)\longrightarrow u(x)$ for all$x\in H$.
I know that $\cdot : B(H)\times B(H) \longrightarrow B(H)$ such that $(u,v)\longmapsto uv$ is separately continuous and jointly continuous on bounded set.
Q:I need some nets or sequences in $B(H)$ such that product is not jointly continuous i.e I need $(u_{\alpha}) $ and$ (v_{\alpha})$ in $B(H)$ such that $u_{\alpha}\longrightarrow u$ and $v_{\alpha}\longrightarrow u$ in S.O.T topology but $u_{\alpha}v_{\alpha}\nrightarrow uv$
Let $I$ be the directed set of finite dimensional subspaces of $H$, ordered with inclusion. Define for such a subspace $u_V = 0\cdot \pi_V +\, 2^{|V|}\cdot\pi_{V^\perp}$, here $\pi$ is the projection onto the subspace in the index and $|V|$ the dimension of $V$. This (unbounded) net converges to zero in the strong topology, as $u_V$ restricted to $V$ is zero.
For a general definition of $v_V$ we need the axiom of choice, but for most specific Hilbert space it is probably possible to do this without choice, we'll see that for $\ell^2(\Bbb N)$, and since for most Hilbert spaces you consider you have explicit isomorphisms to $\ell^2(\Bbb N)$ it works without choice for those too.
For any finite dimensional subspace $V$ choose a norm $1$ element $x_V$ from $V^\perp$. Further choose an arbitrary norm one element of $H$, $x$ (this element does not depend on $V$). Let $v_V(x)=\frac{x_V\otimes x^*}{|V|}$, ie the map that sends $x$ to $x_V$ with scaling $\frac1{|V|}$ and zero on the complement. This net converges to $0$ in norm and thus also in the strong operator topology.
Now $\|u_Vv_V(x)\| = \left\|\frac{2^{|V|}}{|V|}x_V\right\|=\frac{2^{|V|}}{|V|}$, so $u_Vv_V$ does not converge to zero in SOT.
To get rid of choice in $\ell^2(\Bbb N)$, let $(e_n)_{n\in\Bbb N}$ be the standard Hilbert basis, let $x_V$ be the first non-zero $\pi_{V^\perp} e_n$ (rescaled to norm one) and $x$ be $e_1$.